CHAPTER 3 STOICHIOMETRY OF FORMULAS AND EQUATIONS 3.1 Cl 35.45 amu 35.45 g/mol Cl Mass Cl = 35.45 g Cl(2 mol Cl)

1 mol Cl

= 70.90 g Cl Al 26.98 amu 26.98 g/mol Al Mass Al = 26.98 g Al(3 mol Al)

1 mol Al

= 80.94 g Al 3.2 Plan: The formulas are based on the mole ratios of the constituents. Avogadros number allows the change from moles to atoms. Solution:

a) Moles of C atoms = ( )12 22 11

12 mol C1 mol Sucrose1 mol C H O

= 12 mol C

b) C atoms = ( ) 2312 22 1112 22 11

12 mol C 6.022 x 10 C atoms2 molC H O1 mol C H O 1 mol C

= 1.445 x 1025 C atoms

3.3 1 mol of chlorine could be interpreted as a mole of chlorine atoms or a mole of chlorine molecules, Cl2. Specify

which to avoid confusion. The same problem is possible with other diatomic or polyatomic molecules, e.g., F2, Br2, I2, H2, O2, N2, S8, and P4. For these elements, as for chlorine, it is not clear if atoms or molecules are being discussed.

3.4 The molecular mass is the sum of the atomic masses of the atoms or ions in a molecule. The molar mass is the mass of 1 mole of a chemical entity. Both will have the same numeric value for a given chemical substance but molecular mass will have the units of amu and molar mass will have the units of g/mol. 3.5 The mole maintains the same mass relationship between macroscopic samples as exist between individual chemical entities. It relates the number of chemical entities (atoms, molecules, ions, electrons) to the mass. 3.6 Plan: It is possible to relate the relative atomic masses by counting the number of atoms. Solution: a) The element on the left (green) has the higher molar mass because only 5 green balls are necessary to counterbalance the mass of 6 yellow balls. Since the green ball is heavier, its atomic mass is larger, and therefore its molar mass is larger. b) The element on the left (red) has more atoms per gram. This figure requires more thought because the number of red and blue balls is unequal and their masses are unequal. If each pan contained 3 balls, then the red balls would be lighter. The presence of 6 red balls means that they are that much lighter. Because the red ball is lighter, more red atoms are required to make 1 gram. c) The element on the left (orange) has fewer atoms per gram. The orange balls are heavier, and it takes fewer orange balls to make 1 gram. d) Neither element has more atoms per mole. Both the left and right elements have the same number of atoms per mole. The number of atoms per mole (6.022 x 1023) is constant and so is the same for every element. 3.7 Plan: Locate each of the elements on the periodic table and record its atomic mass. The atomic mass of the element times the number of atoms present in the formula gives the mass of that element in one mole of the substance. The molar mass is the sum of the masses of the elements in the substance.

3-1

Solution: a) The molar mass, M, is the sum of the atomic weights, expressed in g/mol: Sr = (1 mol Sr) (87.62 g Sr/mol Sr) = 87.62 g Sr/mol Sr(OH)2 O = (2 mol O) (16.00 g O/mol O) = 32.00 g O/mol Sr(OH)2 H = (2 mol H) (1.008 g H/mol H) = 2.016 g H/mol Sr(OH)2

= 121.64 g/mol of Sr(OH)2 b) M = (2 mol N) (14.01 g N/mol N) + (3 mol O) (16.00 g O/mol O) = 76.02 g/mol of N2O3 c) M = (1 mol Na) (22.99 g Na/mol Na) + (1 mol Cl) (35.45 g Cl/mol Cl) + (3 mol O) (16.00 g O/mol O) = 106.44 g/mol of NaClO3 d) M = (2 mol Cr) (52.00 g Cr/mol Cr) + (3 mol O) (16.00 g O/mol O) = 152.00 g/mol of Cr2O3 3.8 a) (NH4)3PO4 = 3(14.01) + 12(1.008) + 30.97 + 4(16.00) = 149.10 g/mol b) CH2Cl2 = 12.01 + 2(1.008) + 2(35.45) = 84.93 g/mol c) CuSO45H2O = 63.55 + 32.07 + 9(16.00) + 10(1.008) = 249.70 g/mol d) BrF3 = 79.90 + 3(19.00) = 136.90 g/mol 3.9 Plan: Locate each of the elements on the periodic table and record its atomic mass. The atomic mass of the element times the number of atoms present in the formula gives the mass of that element in one mole of the

substance. The molar mass is the sum of the masses of the elements in the substance. Solution: a) M = (1 mol Sn) (118.7 g Sn/mol Sn) + (1 mol O) (16.00 g O/mol O) = 134.7 g/mol of SnO b) M = (1 mol Ba) (137.3 g Ba/mol Ba) + (2 mol F) (19.00 g F/mol F) = 175.3 g/mol of BaF2 c) M = (2 mol Al) (26.98 g Al/mol Al) + (3 mol S) (32.07 g S/mol S) + (12 mol O) (16.00 g O/mol O) = 342.17 g/mol of Al2(SO4)3 d) M = (1 mol Mn) (54.94 g Mn/mol Mn) + (2 mol Cl) (35.45 g Cl/mol Cl) = 125.84 g/mol of MnCl2 3.10 a) N2O4 = 2(14.01) + 4(16.00) = 92.02 g/mol b) C8H10 = 8(12.01) + 10(1.008) = 106.16 g/mol c) MgSO47H2O = 24.31 + 32.07 + 11(16.00) + 14(1.008) = 246.49 g/mol d) Ca(C2H3O2)2 = 40.08 + 4(12.01) + 6(1.008) + 4(16.00) = 158.17 g/mol 3.11 Plan: The mass of a substance and its number of moles are related through the conversion factor of M, the molar mass expressed in g/mol. The moles of a substance and the number of entities per mole are related by the conversion factor, Avogadros number. Solution: a) M of KMnO4 = 39.10 + 54.94 + (4 x 16.00) = 158.04 g/mol of KMnO4

Mass of KMnO4 = ( ) 444

158.04 g KMnO0.68 mol KMnO

1 mol KMnO

= 107.47 = 1.1 x 102 g KMnO4

b) M of Ba(NO3)2 = 137.3 + (2 x 14.01) + (6 x 16.00) = 261.3 g/mol Ba(NO3)2

Moles of O atoms = ( ) 3 23 23 2 3 2

1 mol Ba(NO ) 6 mol O atoms8.18 g Ba(NO )261.3 g Ba(NO ) 1 mol Ba(NO )

= 0.18783 = 0.188 mol O atoms c) M of CaSO4 2H2O = 40.08 + 32.07 + (6 x 16.00) + (4 x 1.008) = 172.18 g/mol (Note that the waters of hydration are included in the molar mass.)

O atoms = ( ) 233 1 mol Ca Cmpd 6 mol O atoms 6.022 x 10 O atoms7.3 x 10 g Ca Cmpd 172.18 g Ca Cmpd 1 mol Ca Cmpd 1 mol O atoms = 1.5319 x 1020 = 1.5 x 1020 O atoms

3-2

3.12 a) Mass NO2 = ( )21 2 22 23 322

1 mol NO 46.01 g NO 1 kg4.6 x10 molecules NO1 mol NO6.022 x 10 molecules NO 10 g

= 3.5145 x 104 = 3.5 x 104 kg NO2

b) Moles Cl atoms = ( ) 2 4 22 4 22 4 2 2 4 2

1 mol C H Cl 2 mol Cl atoms0.0615 g C H Cl98.95 g C H Cl 1 mol C H Cl

= 1.2431 x 103 = 1.24 x 103 mol Cl atoms

c) Number of H ions = ( ) 23222 2

1 mol SrH 2 mol H 6.022 x 10 H ions5.82 g SrH89.64 g SrH 1 mol SrH 1 mol H

= 7.81973 x 1022 = 7.82 x 1022 H ions 3.13 Plan: Determine the molar mass of each substance, then perform the appropriate molar conversions. Solution: a) M of MnSO4 = (54.94 g Mn/mol Mn) + (32.07 g S/mol S) + [(4 mol O) (16.00 g O/mol O)] = 151.01 g/mol of MnSO4

Mass of MnSO4 = ( )2 444

151.01 g MnSO6.44 x 10 mol MnSO

1 mol MnSO

= 9.7250 = 9.73 g MnSO4

b) M of Fe(ClO4)3 = (55.85 g Fe/mol Fe) + [(3 mol Cl) (35.45 g Cl/mol Cl)] + [(12 mol O) (16.00 g O/mol O)] = 354.20 g/mol of Fe(ClO4)3

Moles Fe(ClO4)3 = ( ) 3 4 34 34 3

1 mol Fe(ClO )10 g15.8 kg Fe(ClO )1 kg 354.20 g Fe(ClO )

= 44.6076 = 44.6 mol Fe(ClO4)3 c) M of NH4NO2 = [(2 mol N) (14.01 g N/mol N)] + [(4 mol H) (1.008 g H/mol H)] + [(2 mol O) (16.00 g O/mol O)] = 64.05 g/mol of NH4NO2

N atoms =

( ) 3 234 24 24 2 4 2

1 mol NH NO10 g 2 mol N atoms 6.022 x 10 N atoms92.6 mg NH NO1 mg 64.05 g NH NO 1 mol NH NO 1 mol N atoms

= 1.74126 x 1021 = 1.74 x 1021 N atoms 3.14 FU = formula unit

a) Ions = ( ) 232 222 2

1 mol SrF 6.022 x 10 FU SrF 3 ions38.1 g SrF125.62 g SrF 1 mol SrF 1 FU SrF

2

= 5.47934 x 1023 = 5.48 x 1023 ions

b) Mass CuCl22H2O = ( ) 2 22 2 32 2

170.48 g CuCl 2H O 1 kg3.58 molCuCl 2H O 1 mol CuCl 2H O 10 g

ii i

= 0.61032 = 0.610 kg CuCl22H2O

c) Mass Bi(NO3)35H2O = ( )22 3 3 223 33 3 2

485.1 g Bi(NO ) 5H O1 mol 1 mg2.88 x 10 FU1 FU Bi(NO ) 5H O6.022 x 10 FU 10 g

i

i

= 23199.7 = 2.32 x 104 mg Bi(NO3)35H2O

3-3

3.15 Plan: The formula of each compound must be determined from its name. The molar mass for each formula comes from the formula and atomic masses from the periodic table. Avogadros number is also necessary to find . the number of particles. Solution:

a) Carbonate is a polyatomic anion with the formula, CO32. Copper (I) indicates Cu+. The correct formula for this ionic compound is Cu2CO3.

M of Cu2CO3 = (2 x 63.55) + 12.01 + (3 x 16.00) = 187.11 g/mol

Mass Cu2CO3 = ( ) 2 32 32 3

187.11 g Cu CO8.35 mol Cu CO

1 mol Cu CO

= 1562.4 = 1.56 x 103 g Cu2CO3

b) Dinitrogen pentaoxide has the formula N2O5. Di- indicates 2 N atoms and penta- indicates 5 O atoms. M of N2O5 = (2 x 14.01) + (5 x 16.00) = 108.02 g/mol

Mass N2O5 = ( )20 2 5 2 52 5 232 52 5

1 mol N O 108.02 g N O4.04 x 10 N O molecules

1 mol N O6.022 x 10 N O molecules

= 0.072468 = 0.0725 g N2O5 c) The correct formula for this ionic compound is NaClO4. There are Avogadros number of entities (in this case,

formula units) in a mole of this compound. M of NaClO4 = 22.99 + 35.45 + (4 x 16.00) = 122.44 g/mol

Moles NaClO4 = ( ) 444

1 mol NaClO78.9 g NaClO

122.44 g NaClO

= 0.644397 = 0.644 mol NaClO4

FU = formula units

FU NaClO4 = ( ) 234 444 4

1 mol NaClO 6.022 x 10 FU NaClO78.9 g NaClO

122.44 g NaClO 1 mol NaClO

=

3.88056 x 1023 = 3.88 x 1023 FU NaClO4 d) The number of ions or atoms is calculated from the formula units given in part c. Note the unrounded initially calculated value is used to avoid intermediate rounding.

3.88056 x 1023 FU NaClO4 4

1 Na ion1 FU NaClO

+ = 3.88 x 1023 Na+ ions

3.88056 x 1023 FU NaClO4 44

1 ClO ion1 FU NaClO

= 3.88 x 1023 ClO4 ions

3.88056 x 1023 FU NaClO4 4

1 Cl atom1 FU NaClO

= 3.88 x 1023 Cl atoms

3.88056 x 1023 FU NaClO4 4

4 O atoms1 FU NaClO

= 1.55 x 1024 O atoms

3.16 a) The formula is Cr2(SO4)310H2O, and the molar mass is 572.4 g/mol.

Grams Cr2(SO4)310H2O = ( )2 4 3 2 572.4 g8.42 mol Cr (SO ) 10H O mol = 4819.61 = 4.82 x 103 g Cr2(SO4)310H2O b) The formula is Cl2O7, and the molar mass is 182.9 g/mol.

Grams Cl2O7 = ( )24 2 72 7 23 182.9 g Cl O1 mol1.83 x 10 molecules Cl O 1 mol6.022 x 10 molecules = 555.807 = 5.56 x 102 g Cl2O7

c) The formula is Li2SO4, and the molar mass is 109.95 g/mol.

Mol Li2SO4 = ( ) 2 42 42 4

1 mol Li SO6.2 g Li SO

109.95 g Li SO

= 0.056389 = 0.056 mol Li2SO4

3-4

FU Li2SO4 = ( ) 232 42 42 4 2 4

1 mol Li SO 6.022 x 10 FU6.2 g Li SO109.95 g Li SO 1 mol Li SO

= 3.39576 x 1022 = 3.4 x 1022 FU Li2SO4 d) Note the unrounded initially calculated FU value is used to avoid intermediate rounding.

3.39576 x 1022 FU Li2SO4 2 4

2 Li ions1 FU Li SO

+ = 6.7915 x 1022 = 6.8 x 1022 Li+ ions

3.39576 x 1022 FU Li2SO4 2

4

2 4

1 SO ion1 FU Li SO

= 3.39576 x 1022 = 3.4 x 1022 SO42 ions

3.39576 x 1022 FU Li2SO4 2 4

1 S atom1 FU Li SO

= 3.39576 x 1022 = 3.4 x 1022 S atoms

3.39576 x 1022 FU Li2SO4 2 4

4 O atoms1 FU Li SO

= 1.3583 x 1023 = 1.4 x 1023 O atoms

3.17 Plan: Determine the formula and the molar mass of each compound. The formula gives the number of atoms of each type of element present. Masses come from the periodic table. Mass percent = (total mass of element in the substance/molar mass of substance) x 100. Solution: a) Ammonium bicarbonate is an ionic compound consisting of ammonium ions, NH4+ and bicarbonate ions, HCO3. The formula of the compound is NH4HCO3. M of NH4HCO3 = (14.01 g/mol) + (5 x 1.008 g/mol) + (12.01 g/mol) + (3 x 16.00 g/mol) = 79.06 g/mol NH4HCO3 In 1 mole of ammonium bicarbonate, with a mass of 79.06 g, there are 5 H atoms with a mass of 5.040 g.

4 3

(5 mol H) (1.008 g/mol H) x 100%79.06 g/mol NH HCO

= 6.374905 = 6.375% H

b) Sodium dihydrogen phosphate heptahydrate is a salt that consists of sodium ions, Na+, dihydrogen phosphate ions, H2PO4, and seven waters of hydration. The formula is NaH2PO47H2O. Note that the waters of hydration are included in the molar mass. M of NaH2PO47H2O = (22.99 g/mol) + (16 x 1.008 g/mol) + (30.97g/mol) + (11 x 16.00 g/mol) = 246.09 g/mol NaH2PO47H2O In each mole of NaH2PO47H2O (with mass of 246.09 g), there are 11 x 16.00 g/mol or 176.00 g of oxygen.

( )( )11 mol O 16.00 g/mol O

x 100%246.09 g/mol

= 71.51855 = 71.52% O

3.18 a) Sr(IO4)2 469.4 g/mol

( )( )2 mol I 126.9 g/mol I

x 100%469.4 g/mol

= 54.0690 = 54.07% I

b) KMnO4 158.04 g/mol

( )( )1 mol Mn 54.94 g/mol Mn

x 100%158.04 g/mol

= 34.76335 = 34.76% Mn

3-5

3.19 Plan: Determine the formula of cisplatin from the figure, and then calculate the molar mass from the formula. The molar mass is necessary for the subsequent calculations. Solution: The formula for cisplatin is Pt(Cl)2(NH3) 2 M of Pt(Cl)2(NH3) 2 = 195.1 + (2 x 35.45) + (2 x 14.01) + (6 x 1.008) = 300.1 g/mol

a) Moles cisplatin = ( ) 1 mol cisplatin285.3 g cisplatin300.1 g cisplatin

= 0.9506831 = 0.9507 mol cisplatin

b) H atoms = ( ) 236 mol H 6.022 x 10 H atoms0.98 mol cisplatin1 mol cisplatin 1 mol H

= 3.540936 x 1024 = 3.5 x 1024 H atoms

3.20 a) Moles C3H8 = ( ) 3 83 83 8

1 mol C H85.5 g C H

44.09 g C H

= 1.939215 = 1.94 mol C3H8

b) Grams C = ( ) 3 83 83 8 3 8

1 mol C H 3 mol C 12.01 g C85.5 g C H44.09 g C H 1 mol C H 1 mol C

= 69.869925 = 69.9 g C

3.21 Plan: Determine the formulas for the compounds where needed. Determine the molar mass of each formula. Calculate the percent nitrogen by dividing the mass of nitrogen in a mole of compound by the molar mass of the compound, and multiply the result by 100%. Then rank the values. Solution: Name Formula Molar Mass (g/mol) Potassium nitrate KNO3 101.11 Ammonium nitrate NH4NO3 80.05 Ammonium sulfate (NH4)2SO4 132.15 Urea CO(NH2)2 60.06 Calculating the nitrogen percentages:

Potassium nitrate ( ) ( )1 mol N 14.01 g/mol N x 100101.11 g/mol

= 13.856196 = 13.86% N

Ammonium nitrate ( )( )2 mol N 14.01 g/mol N x 10080.05 g/mol

= 35.003123 = 35.00% N

Ammonium sulfate ( )( )

100 x g/mol 132.15

N g/mol 14.01 N mol 2 = 21.20318 = 21.20% N

Urea ( )( )2 mol N 14.01 g/mol N

x 10060.06 g/mol

= 46.6533 = 46.65% N

Rank is CO(NH2)2 > NH4NO3 > (NH4)2SO4 > KNO3 3.22 Plan: The volume must be converted from cubic feet to cubic centimeters (or vice versa). The volume and the

density will give you mass, and the mass with the molar mass gives you moles. Part (b) requires a conversion from cubic decimeters, instead of cubic feet, to cubic centimeters. The density allows you to change these cubic centimeters to mass, the molar mass allows you to find moles, and finally Avogadros number allows you to make the last step. Solution: The molar mass of galena (PbS) is 239.3 g/mol.

a) Moles PbS = ( )

PbS g 239.3PbS mol 1

cm 1PbS g 46.7

in) 1(cm) 54.2(

ft) (1in) 12(

PbS ft 00.133

3

3

33

= 882.7566886 = 883 mol PbS

3-6

b) Lead atoms =

( ) 3 3 233 3 2 3 3(0.1 m) (1 cm) 7.46 g PbS 1 mol PbS 1 mol Pb 6.022 x 10 Pb atoms1.00 dm PbS 239.3 g PbS 1 mol PbS 1 mol Pb(1 dm) (10 m) 1 cm

= 1.87731 x 1025 = 1.88 x 1025 Pb atoms 3.23 If the molecular formula for hemoglobin (Hb) were known, the number of Fe2+ ions in a molecule of hemoglobin could be calculated. It is possible to calculate the mass of iron from the percentage of iron and the molar mass of the compound. From the mass of iron, the moles of iron per mole of hemoglobin may be found.

40.33% Fe 6.8 x 10 g 1 mol Fe

100% Hb mol 55.85 g Fe

= 4.0179 = 4.0 mol Fe2+/mol Hb

Thus, there are 4 Fe2+/molecule Hb. 3.24 Plan: Remember that the molecular formula tells the actual number of moles of each element in one mole of

compound. Solution: a) No, this information does not allow you to obtain the molecular formula. You can obtain the empirical formula from the number of moles of each type of atom in a compound, but not the molecular formula.

b) Yes, you can obtain the molecular formula from the mass percentages and the total number of atoms. Solution plan:

1) Assume a 100.0 g sample and convert masses (from the mass % of each element) to moles using molar mass.

2) Identify the element with the lowest number of moles and use this number to divide into the number of moles for each element. You now have at least one elemental mole ratio (the one with the smallest number of moles) equal to 1.00 and the remaining mole ratios that are larger than one.

3) Examine the numbers to determine if they are whole numbers. If not, multiply each number by a whole number factor to get whole numbers for each element. You will have to use some judgment to decide when to round.

4) Write the empirical formula using the whole numbers from step 3. 5) Check the total number of atoms in the empirical formula. If it equals the total number of atoms given then the empirical formula is also the molecular formula. If not, then divide the total number of atoms given by the total number of atoms in the empirical formula. This should give a whole number. Multiply the number of atoms of each element in the empirical formula by this whole number to get the molecular formula. If you do not get a whole number when you divide, return to step 3 and revise how you multiplied and rounded to get whole numbers for each element. c) Yes, you can determine the molecular formula from the mass percent and the number of atoms of one element in a compound. Solution plan: 1) Follow steps 1 4 in part b. 2) Compare the number of atoms given for the one element to the number in the empirical formula. Determine the factor the number in the empirical formula must be multiplied by to obtain the given number of atoms for that element. Multiply the empirical formula by this number to get the molecular formula. d) No, the mass % will only lead to the empirical formula. e) Yes, a structural formula shows all the atoms in the compound. Solution plan: Count the number of atoms of each type of element and record as the number for the molecular formula. 3.25 Plan: Examine the number of atoms of each type in the compound. Divide all atom numbers by any common factor. The final answers must be the lowest whole-number values. Solution: a) C2H4 has a ratio of 2 carbon atoms to 4 hydrogen atoms, or 2:4. This ratio can be reduced to 1:2, so that the empirical formula is CH2. The empirical formula mass is 12.01 + 2(1.008) = 14.03 g/mol. b) The ratio of atoms is 2:6:2, or 1:3:1. The empirical formula is CH3O and its empirical formula mass is 12.01 + 3(1.008) + 16.00 = 31.03 g/mol.

3-7

c) Since, the ratio of elements cannot be further reduced, the molecular formula and empirical formula are the same, N2O5. The formula mass is 2(14.01) + 5(16.00) = 108.02 g/mol. d) The ratio of elements is 3 atoms of barium to 2 atoms of phosphorus to 8 atoms of oxygen, or 3:2:8. This ratio cannot be further reduced, so the empirical formula is also Ba3(PO4)2. The formula mass is 3(137.3) + 2(30.97) + 8(16.00) = 601.8 g/mol.

e) The ratio of atoms is 4:16, or 1:4. The empirical formula is TeI4, and the formula mass is 127.6 + 4(126.9) = 635.2 g/mol.

3.26 a) C4H8 = M.F. CH2 = E.F. E.F. mass = 14.03 g/mol b) C3H6O3 = M.F. CH2O = E.F. E.F. mass = 30.03 g/mol c) P4O10 = M.F. P2O5 = E.F. E.F. mass = 141.94 g/mol d) Ga2(SO4)3 = M.F. Ga2(SO4)3 = M.F. = E.F. E.F. mass = 427.6 g/mol e) Al2Br6 = M.F. AlBr3 = E.F. E.F. mass = 266.7 g/mol 3.27 Plan: Determine the molar mass of each empirical formula. The molar mass of each compound divided by its empirical formula mass gives the number of times the empirical formula is within the molecule. Multiply the empirical formula by the number of times the empirical formula appears to get the molecular formula.

Solution: Only approximate whole number values are needed. a) CH2 has empirical mass equal to 14.03 g/mol

42.08 g/mol14.03 g/mol

= 3

Multiplying the subscripts in CH2 by 3 gives C3H6 b) NH2 has empirical mass equal to 16.03 g/mol

32.05 g/mol16.03 g/mol

= 2

Multiplying the subscripts in NH2 by 2 gives N2H4 c) NO2 has empirical mass equal to 46.01 g/mol

92.02 g/mol46.01 g/mol

= 2

Multiplying the subscripts in NO2 by 2 gives N2O4 d) CHN has empirical mass equal to 27.03 g/mol

135.14 g/mol27.03 g/mol

= 5

Multiplying the subscripts in CHN by 5 gives C5H5N5 3.28 a) CH has empirical mass equal to = 13.02 g/mol

78.11 g/mol13.02 g/mol

= 6 M.F. = 6 x E.F. = C6H6

b) C3H6O2 has empirical mass equal to = 74.08 g/mol

74.08 g/mol74.08 g/mol

= 1 M.F. = 1 x E.F. = C3H6O2

c) HgCl has empirical mass equal to = 236.0 g/mol

472.1 g/mol236.0 g/mol

= 2 M.F. = 2 x E.F. = Hg2Cl2

3-8

d) C7H4O2 has empirical mass equal to = 120.10 g/mol

240.20 g/mol120.10 g/mol

= 2 M.F. = 2 x E.F. = C14H8O4

3.29 Plan: The empirical formula is the smallest whole-number ratio of the atoms or moles in a formula. All data must be converted to moles of an element. Using the smallest number of moles present, convert the mole ratios to whole numbers. Solution:

a) 0.063 mol Cl0.063 mol Cl

= 1 0.22 mol O0.063 mol Cl

= 3.5

The formula is Cl1O3.5, which in whole numbers (x 2) is Cl2O7.

b) ( ) 1 mol Si2.45 g Si28.09 g Si

= 0.08722 mol Si ( ) 1 mol Cl12.4 g Cl

35.45 g Cl

= 0.349788 mol Cl

0.08722 mol Si0.08722 mol Si

= 1 0.349788 mol Cl

0.08722 mol Si

= 4

The empirical formula is SiCl4. c) Assume a 100 g sample and convert the masses to moles.

( ) 27.3% C 1 mol C100 g100% 12.01 g C

= 2.2731 mol C ( ) 72.7% O 1 mol O100 g

100% 16.00 g O

= 4.5438 mol O

2.2731 mol C2.2731 mol C

= 1 4.5438 mol O

2.2731 mol C

= 2

The empirical formula is CO2.

3.30 a) 0.039 mol Fe0.039 mol Fe

= 1 0.052 mol O

0.039 mol Fe

= 1.3333

The formula is Fe1O1.3333, which in whole numbers (x 3) is Fe3O4.

b) ( ) 1 mol P0.903 g P30.97 g P

= 0.029157 mol P ( ) 1 mol Br6.99 g Br 79.90 g Br

= 0.087484 mol Br

0.029157 mol P0.029157 mol P

= 1 0.087484 mol Br

0.029157 mol P

= 3

The empirical formula is PBr3. c) Assume a 100 g sample and convert the masses to moles.

( ) 79.9% C 1 mol C100 g100% 12.01 g C

= 6.6528 mol C

( ) (100 79.9)% H 1 mol H100 g100% 1.008 g H

= 19.940 mol H

6.6528 mol C6.6528 mol C

= 1 19.940 mol H

6.6528 mol C

= 3

The empirical formula is CH3.

3-9

3.31 Plan: The balanced equation for this reaction is: M(s) + F2(g) MF2(s) since fluorine, like other halogens, exists as a diatomic molecule. The moles of the metal are known, and the moles of everything else may be found from these moles using the balanced chemical equation. Solution: a) Determine the moles of fluorine.

Moles F = ( ) 2 mol F0.600 mol M1 mol M

= 1.20 mol F b) The grams of M are the grams of MF2 minus the grams of F present.

Grams M = 46.8 g (M + F) ( ) 19.00 g F1.20 mol F1 mol F

= 24.0 g M

c) The molar mass is needed to identify the element. Molar mass of M = 24.0 g M/0.600 mol M = 40.0 g/mol The metal with the closest molar mass to 40.0 g/mol is calcium.

3.32 a) Moles O = ( )2 32 3

3 mol O0.370 mol M O1 mol M O

= 1.11 mol O

b) The grams of M are the grams of M2O3 minus the grams of O present.

Grams M = 55.4 g (M + O) ( ) 16.00 g O1.11 mol O1 mol O

= 37.64 = 37.6 g M

c) First, the number of moles of M must be calculated.

Moles M = ( )2 32 3

2 mol M0.370 mol M O1 mol M O

= 0.740 mol M

The molar mass is needed to identify the element. (Use the unrounded mass of M to avoid intermittent rounding errors.) Molar mass of M = 37.64 g M/0.740 mol M = 50.86 g/mol The metal with the closest molar mass to 50.9 g/mol is vanadium. 3.33 Plan: Assume 100 grams of cortisol so the percentages are numerically equivalent to the masses of each element. Convert each of the masses to moles by using the molar mass of each element involved. Divide all moles by the lowest number of moles and convert to whole numbers to determine the empirical formula. The empirical formula mass and the given molar mass will then relate the empirical formula to the molecular formula. Solution:

Moles C = ( ) 1 mol C69.6 g C12.01 g C

= 5.7952 mol C

Moles H = ( ) 1 mol H8.34 g H1.008 g H

= 8.2738 mol H

Moles O = ( ) 1 mol O22.1 g O16.00 g O

= 1.38125 mol O

5.7952 mol C1.38125 mol O

= 4.20 8.2738 mol H1.38125 mol O

= 6.00 1.38125 mol O1.38125 mol O

= 1.00

The carbon value is not close enough to a whole number to round the value. The smallest number that 4.20 may be multiplied by to get close to a whole number is 5. (You may wish to prove this to yourself.) All three ratios need to be multiplied by five to get the empirical formula of C21H30O5.

The empirical formula mass is: 21 (12.01 g C/mol) + 30 (1.008 g H/mol) + 5 (16.00 g O/mol) = 362.45 g/mol The empirical formula mass and the molar mass given are the same, so the empirical and the molecular formulas are the same. The molecular formula is C21H30O5.

3-10

3.34 Plan: In combustion analysis, finding the moles of carbon and hydrogen is relatively simple because all of the carbon present in the sample is found in the carbon of CO2, and all of the hydrogen present in the sample is found in the hydrogen of H2O. The moles of oxygen are more difficult to find, because additional O2 was added to cause the combustion reaction. The masses of CO2 and H2O are used to find both the mass of C and H and the moles of C and H. Subtracting the masses of C and H from the mass of the sample gives the mass of O. Convert the mass of O to moles of O. Take the moles of C, H, and O and divide by the smallest value, and convert to a whole number to get the empirical formula. Determine the empirical formula mass and compare it to the molar mass given in the problem to see how the empirical and molecular formulas are related. Finally, determine the molecular formula.

Solution: (There is no intermediate rounding.) Initial mole determination:

Moles C = ( ) 222 2

1 mol CO 1 mol C0.449 g CO44.01 g CO 1 mol CO

= 0.010202 mol C

Moles H = ( ) 222 2

1 mol H O 2 mol H0.184 g H O18.02 g H O 1 mol H O

= 0.020422 mol H

Now determine the masses of C and H:

Grams C = ( ) 12.01 g C0.010202 mol C1 mol C

= 0.122526 g C

Grams H = ( ) 1.008 g H0.020422 mol H1 mol H

= 0.020585 g H

Determine the mass and then the moles of O: 0.1595 g (C, H, and O) (0.122526 g C + 0.020585 g H) = 0.016389 g O

Moles O = ( ) 1 mol O0.016389 g O16.00 g O

= 0.0010243 mol O Divide by the smallest number of moles: (Rounding is acceptable for these answers.)

0.010202 mol C0.0010243 mol O

= 9.96 = 10 0.020422 mol H

0.0010243 mol O

= 19.9 = 20 0.0010243 mol O0.0010243 mol O

= 1

Empirical formula = C10H20O Empirical formula mass = 10 (12.01 g C/mol) + 20 (1.008 g H/mol) + 1 (16.00 g O/mol) = 156.26 g/mol The empirical formula mass matches the given molar mass so the empirical and molecular formulas are the same. The molecular formula is C10H20O. 3.35 Students I and II are incorrect. Both students changed a given formula. Only coefficients should be changed when balancing. Student I failed to identify the product correctly, and Student II used atomic chlorine instead of molecular chlorine as a reactant. Student III followed the correct process. 3.36 Plan: Examine the diagram and label each formula. We will use A for red atoms and B for green atoms. Solution: The reaction shows A2 and B2 molecules forming AB molecules. Equal numbers of A2 and B2 combine to give twice as many molecules of AB. Thus, the reaction is A2 + B2 2 AB. This is the answer to part b. 3.37 Plan: Balancing is a trial and error procedure. Do one blank/one element at a time. Solution:

a) 16 Cu(s) + __ S8(s) 8 Cu2S(s) Hint: Balance the S first, because there is an obvious deficiency of S on the right side of the equation. Then balance the Cu.

b) __ P4O10(s) + 6 H2O(l) 4 H3PO4(l) Hint: Balance the P first, because there is an obvious deficiency of P on the right side of the equation. Balance the H next, because H is present in only one reactant and only one product. Balance the O last, because it appears in both reactants and is harder to balance.

3-11

c) __ B2O3(s) + 6 NaOH(aq) 2 Na3BO3(aq) + 3 H2O(l) Hint: Oxygen is again the hardest element to balance because it is present in more than one place on each side of the reaction. If you balance the easier elements first (B, Na, H), the oxygen will automatically be balanced. d) 2 CH3NH2(g) + 9/2 O2(g) 2 CO2(g) + 5 H2O(g) + __ N2(g) 4 CH3NH2(g) + 9 O2(g) 4 CO2(g) + 10 H2O(g) + 2 N2(g) Hint: You should balance odd/even numbers of oxygen using the half method, and then multiply all coefficients by two. 3.38 a) Cu(NO3)2(aq) + 2 KOH(aq) Cu(OH)2(s) + 2 KNO3(aq) b) BCl3(g) + 3 H2O(l) H3BO3(s) + 3 HCl(g) c) CaSiO3(s) + 6 HF(g) SiF4(g) + CaF2(s) + 3 H2O(l) d) (CN)2(g) + 4 H2O(l) H2C2O4(aq) + 2 NH3(g) 3.39 Plan: The names must first be converted to chemical formulas. The balancing is a trial and error procedure. Do one blank/one element at a time. Remember that oxygen is diatomic. Solution: a) 4 Ga(s) + 3 O2(g) 2 Ga2O3(s) b) 2 C6H14(l) + 19 O2(g) 12 CO2(g) + 14 H2O(g) c) 3 CaCl2(aq) + 2 Na3PO4(aq) Ca3(PO4)2(s) + 6 NaCl(aq) 3.40 a) Pb(NO3)2(aq) + 2 KI(aq) PbI2(s) + 2 KNO3(aq) b) Si2Cl6(l) + 4 H2O(l) 2 SiO2(s) + 6 HCl(g) + H2(g) c) 3 NO2(g) + H2O(l) 2 HNO3(aq) + NO(g) 3.41 a) The reaction is A2 + 3 B2 2 AB3 so the mole ratio between A2 and B2 is 1:3. Three times as many B2 molecules are required as you have of A2 molecules. With 3 A2 molecules present, 3 x 3 = 9 B2 molecules would be required. Since you have only 6 B2 molecules, B2 is the limiting reagent.

b) Use the mole ratio between B2 and AB3: ( ) 322

2 AB molecules6 B molecules

3 B molecules

= 4 AB3 molecules 3.42 Plan: Convert the kilograms of oxygen to the moles of oxygen. Use the moles of oxygen and the mole ratios from the balanced chemical equation to determine the moles of KNO3. The moles of KNO3 and its molar mass will give the grams. Solution:

a) Moles KNO3 = ( ) 3 3222 2

4 mol KNO1 mol O10 g56.6 kg O1 kg 32.00 g O 5 mol O

= 1415 = 1.42 x 103 mol KNO3

b) Grams KNO3 = ( ) 3 3 3222 2

4 mol KNO 101.11 g KNO1 mol O10 g56.6 kg O1 kg 32.00 g O 5 mol O 1 mol KNO

3

= 143070.65 = 1.43 x 105 g KNO3 The beginning of the calculation is repeated to emphasize that the second part of the problem is simply an extension of the first part. There is no need to repeat the entire calculation, as only the final step times the answer of the first part will give the final answer to this part.

3.43 a) Moles Cr2O3 = ( ) 2 3 2 32 32 3 2 3

1 mol Cr S 1 mol Cr O421 g Cr S

200.21 g Cr S 1 mol Cr S

= 2.10279 = 2.10 mol Cr2O3

b) Grams Cr2O3 = ( ) 2 3 2 3 2 32 32 3 2 3 2 3

1 mol Cr S 1 mol Cr O 152.00 g Cr O421 g Cr S

200.21 g Cr S 1 mol Cr S 1 mol Cr O

= 319.624 = 3.20 x 102 g Cr2O3

3-12

3.44 Plan: First, balance the equation. Convert the grams of diborane to moles of diborane using its molar mass. Use mole ratios from the balanced chemical equation to determine the moles of the products. Use the moles and molar mass of each product to determine the mass formed. Solution: The balanced equation is: B2H6(g) + 6 H2O(l) 2 H3BO3(s) + 6 H2(g). Mass H3BO3 = ( ) 2 6 3 3 3 32 6

2 6 2 6 3 3

1 mol B H 2 mol H BO 61.83 g H BO43.82 g B H

27.67 g B H 1 mol B H 1 mol H BO

= 195.83597 = 195.8 g H3BO3

Mass H2 = ( ) 2 6 2 22 62 6 2 6 2

1 mol B H 6 mol H 2.016 g H43.82 g B H

27.67 g B H 1 mol B H 1 mol H

= 19.156007 = 19.16 g H2 3.45 First, balance the equation: Ag2S(s) + 2 HCl(aq) 2 AgCl(s) + H2S(g) Grams AgCl = 22

2 2

1 mol Ag S 2 mol AgCl 143.4 g AgCl174g Ag S247.9 g Ag S 1 mol Ag S 1 mol AgCl

= 201.304 = 201 g AgCl

Grams H2S = 2 222 2

1 mol Ag S 1 mol H S 34.09 g H S174 g Ag S

247.9 g Ag S 1 mol Ag S 1 mol H S

2

2

= 23.9276 = 23.9 g H2S

3.46 Plan: Write the balanced equation by first writing the formulas for the reactants and products. Reactants: formula for phosphorus is given as P4 and formula for chlorine gas is Cl2 (chlorine occurs as a diatomic molecule). Products: formula for phosphorus pentachloride the name indicates one phosphorus atom and five chlorine atoms to give the formula PCl5. Convert the mass of phosphorus to moles, use the mole ratio from the balanced chemical equation, and finally use the molar mass of chlorine to get the mass of chlorine. Solution: Formulas give the equation: P4 + Cl2 PCl5 Balancing the equation: P4 + 10 Cl2 4 PCl5 Grams Cl2 = ( ) 4 24

4 4

1 mol P 10 mol Cl 70.90 g Cl455 g P

123.88 g P 1 mol P 1 mol Cl

2

2

= 2604.09267 = 2.60 x 103 g Cl2

3.47 First, balance the equation: S8(s) + 24 F2(g) 8 SF6(s) Grams F2 = ( ) 8 28

8 8

1 mol S 24 mol F 38.00 g F17.8 g S

256.56 g S 1 mol S 1 mol F

2

2 = 63.27409 = 63.3 g F2

3.48 Plan: Convert the given masses to moles and use the mole ratio from the balanced chemical equation to find the moles of CaO that will form. The reactant that produces the least moles of CaO is the limiting reactant. Convert the moles of CaO from the limiting reactant to grams using the molar mass. Solution:

a) Moles CaO from Ca = ( ) 1 mol Ca 2 mol CaO4.20 g Ca40.08 g Ca 2 mol Ca

= 0.104790 = 0.105 mol CaO

b) Moles CaO from O2 = ( ) 222 2

1 mol O 2 mol CaO2.80 g O32.00 g O 1 mol O

= 0.17500 = 0.175 mol CaO

c) Calcium is the limiting reactant since it will form less calcium oxide. d) The mass of CaO formed is determined by the limiting reactant, Ca.

Grams CaO = ( ) 1 mol Ca 2 mol CaO 56.08 g CaO4.20 g Ca40.08 g Ca 2 mol Ca 1 mol CaO

= 5.8766 = 5.88 g CaO

3-13

3.49 SrH2(s) + 2 H2O(l) Sr(OH)2(s) + 2 H2(g) a) Moles H2 from SrH2 = ( ) 22

2 2

1 mol SrH 2 mol H5.70 g SrH

89.64 g SrH 1 mol SrH

2 = 0.127175 = 0.127 mol H2

b) Moles H2 from H2O = ( ) 222 2

1 mol H O 2 mol H4.75 g H O

18.02 g H O 2 mol H O

2 = 0.263596 = 0.264 mol H2

c) SrH2 is the limiting reagent since it will yield fewer moles of hydrogen gas.

d) Grams H2 = ( ) 2 222 2

1 mol SrH 2 mol H 2.016 g H5.70 g SrH

89.64 g SrH 1 mol SrH 1 mol H

2

2

= 0.256386 = 0.256 g H2

3.50 Plan: First, balance the chemical equation. Determine which of the reactants is the limiting reagent. Use the limiting reagent and the mole ratio from the balanced chemical equation to determine the amount of material formed and the amount of the other reactant used. The difference between the amount of reactant used and the initial reactant supplied gives the amount of excess reactant remaining. Solution: The balanced chemical equation for this reaction is: 2 ICl3 + 3 H2O ICl + HIO3 + 5 HCl Hint: Balance the equation by starting with oxygen. The other elements are in multiple reactants and/or products and are harder to balance initially. Next, find the limiting reactant by using the molar ratio to find the smaller number of moles of HIO3 that can be produced from each reactant given and excess of the other:

Moles HIO3 from ICl3 = ( ) 333 3

1 mol ICl 1 mol HIO635 g ICl

233.2 g ICl 2 mol ICl

3 = 1.361492 = 1.36 mol HIO3

Moles HIO3 from H2O = ( ) 3222 2

1 mol HIO1 mol H O118.5 g H O

18.02 g H O 3 mol H O

= 2.19201 = 2.19 mol HIO3

ICl3 is the limiting reagent and will produce 1.36 mol HIO3. Use the limiting reagent to find the grams of HIO3 formed.

Grams HIO3 = ( ) 3 333 3

1 mol ICl 1 mol HIO 175.9 g HIO635 g ICl

233.2 g ICl 2 mol ICl 1 mol HIO

3

3

= 239.4865 = 239 g HIO3

The remaining mass of the excess reagent can be calculated from the amount of H2O combining with the limiting reagent.

Grams H2O required to react with 635 g ICl3: ( ) 3 2 233 3

1 mol ICl 3 mol H O 18.02 g H O635 g ICl

233.2 g ICl 2 mol ICl 1 mol H O 2

= 73.6023 = 73.6 g H2O reacted Remaining H2O = 118.5 g 73.6 g = 44.9 g H2O 3.51 First balance the equation: Al2S3 + 6 H2O(l) 2 Al(OH)3(s) + 3 H2S(g) Determine the limiting reactant:

Moles H2S from Al2S3 = ( ) 2 3 22 32 3 2 3

1 mol Al S 3 mol H S158 g Al S

150.17 g Al S 1 mol Al S

= 3.15642 = 3.16 mol H2S

Moles H2S from H2O = ( ) 222 2

1 mol H O 3 mol H S131 g H O

18.02 g H O 6 mol H O

2 = 3.63485 = 3.63 mol H2S

Al2S3 is the limiting reagent and 3.16 mol of H2S will form. Using the limiting reagent, we can calculate the mass of H2S that will form.

Grams H2S = ( ) 2 3 22 32 3 2 3 2

1 mol Al S 3 mol H S 34.09 g H S158 g Al S

150.17 g Al S 1 mol Al S 1 mol H S

2 = 107.602 = 108 g H2S

3-14

The remaining mass of the excess reagent can be calculated from the amount of H2O combining with the limiting reagent.

Remaining H2O = 131 g H2O ( ) 2 3 2 22 32 3 2 3 2

1 mol Al S 6 mol H O 18.02 g H O158 g Al S

150.17 g Al S 1 mol Al S 1 mol H O

= 17.24 = 17 g H2O 3.52 Plan: Write the balanced equation: formula for carbon is C, formula for oxygen is O2 and formula for carbon dioxide is CO2. Determine the limiting reagent by seeing which reactant will yield the smaller amount of product. The limiting reactant is used for all subsequent calculations. Solution: C(s) + O2(g) CO2(g) Moles CO2 from C = ( ) 21 mol CO0.100 mol C 1 mol C

= 0.100 mol CO2

Moles CO2 from O2 = ( ) 222 2

1 mol O 1 mol CO8.00 g O

32.00 g O 1 mol O

2 = 0.25000 = 0.250 mol CO2

The C is the limiting reactant and will be used to determine the amount of CO2 that will form.

Grams CO2 = ( ) 22

1 mol CO 44.01 g CO0.100 mol C

1 mol C 1 mol CO

2 = 4.401 = 4.40 g CO2 Since the C is limiting, the O2 is in excess. The amount remaining depends on how much combines with the limiting reagent.

O2 required to react with 0.100 mol of C = ( ) 22

1 mol O 32.00 g O0.100 mol C

1 mol C 1 mol O

2 = 3.20 g O2 Remaining O2 = 8.00 g 3.20 g = 4.80 g O2 3.53 First, balance the equation: 2 H2(g) + O2(g) 2 H2O(l) Determine the limiting reagent:

Mole H2O from H2 = ( ) 222 2

1 mol H 2 mol H O0.0375 g H

2.016 g H 2 mol H

2 = 0.018601 = 0.0186 mol H2O

Mole H2O from O2 = ( ) 222

2 mol H O0.0185 mol O

1 mol O

= 0.0370 mol H2O

The hydrogen is the limiting reactant, and will be used to determine the amount of water that will form.

Grams H2O = ( ) 2 222 2

1 mol H 2 mol H O 18.02 g H O0.0375 g H

2.016 g H 2 mol H 1 mol H O

2

2

= 0.335193 = 0.335 g H2O

Since the hydrogen is limiting; the oxygen must be excess reactant. The amount of excess reactant is determined from the limiting reactant.

Remaining O2 = ( ) 222

32.00 mol O0.0185 mol O

1 mol O

( ) 2 222 2

1 mol H 1 mol O 32.00 g O0.0375 g H

2.016 g H 2 mol H 1 mol O

2

2

= 0.294381 = 0.294 g O2 3.54 Plan: The question asks for the mass of each substance present at the end of the reaction. Substance refers to

both reactants and products. Solve this problem using multiple steps. Recognizing that this is a limiting reactant problem, first write a balanced chemical equation. Using the molar relationships from the balanced equation, determine which reactant is limiting. Any product can be used to predict the limiting reactant; in this case, AlCl3 is used. Additional significant figures are retained until the last step. Solution:

The balanced chemical equation is: Al(NO2)3(aq) + 3 NH4Cl(aq) AlCl3(aq) + 3 N2(g) + 6 H2O(l) Now determine the limiting reagent. We will use the moles of AlCl3 produced to determine which is limiting.

3-15

Mole AlCl3 from Al(NO2)3 = ( ) 2 3 32 32 3 2 3

1 mol Al(NO ) 1 mol AlCl72.5 g Al(NO )

165.01 g Al(NO ) 1 mol Al(NO )

= 0.439367 = 0.439 mol AlCl3

Mole AlCl3 from NH4Cl = ( ) 3444 4

1 mol AlCl1 mol NH Cl58.6 g NH Cl

53.49 g NH Cl 3 mol NH Cl

= 0.365177 = 0.365 mol AlCl3

Ammonium chloride is the limiting reactant, and it is used for all subsequent calculations. Mass of substances after the reaction: Al(NO2)3: Al(NO2)3 required to react with 58.6 g of NH4Cl:

( ) 2 3 2 3444 4

1 mol Al(NO ) 165.01 g Al(NO )1 mol NH Cl58.6 g NH Cl

53.49 g NH Cl 3 mol NH Cl 1 mol Al(NO ) 2 3

= 60.2579 = 60.3 g Al(NO2)3 Al(NO2)3 remaining: 72.5 g 60.3 g = 12.2 g Al(NO2)3 NH4Cl: None left since it is the limiting reagent. AlCl3:

( ) 3444 4

1 mol AlCl 133.33 g AlCl1 mol NH Cl58.6 g NH Cl

53.49 g NH Cl 3 mol NH Cl 1 mol AlCl

3

3 = 48.6891 = 48.7 g AlCl3

N2:

( ) 4 244 4

1 mol NH Cl 3 mol N 28.02 g N58.6 g NH Cl

53.49 g NH Cl 3 mol NH Cl 1 mol N

2

2

= 30.697 = 30.7 g N2

H2O:

( ) 4 244 4

1 mol NH Cl 6 mol H O 18.02 g H O58.6 g NH Cl

53.49 g NH Cl 3 mol NH Cl 1 mol H O

2

2

= 39.48297 = 39.5 g H2O

3.55 The balanced chemical equation is: Ca(NO3)2(s) + 2 NH4F(s) CaF2(s) + 2 N2O(g) + 4 H2O(g) Now determine the limiting reagent. We will use the moles of CaF2 produced to determine which is limiting.

Mole CaF2 from Ca(NO3)2 = ( ) 3 2 23 23 2 3 2

1 mol Ca(NO ) 1 mol CaF16.8 g Ca(NO )

164.10 g Ca(NO ) 1 mol Ca(NO )

= 0.1023766 = 0.102 mol CaF2

Mole CaF2 from NH4F = ( ) 444 4

1 mol NH F 1 mol CaF17.50 g NH F

37.04 g NH F 2 mol NH F

2 = 0.236231 = 0.236 mol CaF2

Calcium nitrate is the limiting reactant, and it is important for all subsequent calculations. Mass of substances after the reaction: Ca(NO3)2: None (It is the limiting reactant.) NH4F:

17.50 g NH4F ( ) 3 2 4 43 23 2 3 2 4

1 mol Ca(NO ) 2 mol NH F 37.04 g NH F16.8 g Ca(NO )

164.10 g Ca(NO ) 1 mol Ca(NO ) 1 mol NH F

= 9.9159 = 9.92 g NH4F CaF2:

( ) 3 2 23 23 2 3 2 2

1 mol Ca(NO ) 1 mol CaF 78.08 g CaF16.8 g Ca(NO )

164.10 g Ca(NO ) 1 mol Ca(NO ) 1 mol CaF

2 = 7.99356 = 7.99 g CaF2 N2O:

( ) 3 2 23 23 2 3 2 2

1 mol Ca(NO ) 2 mol N O 44.02 g N O16.8 g Ca(NO )

164.10 g Ca(NO ) 1 mol Ca(NO ) 1 mol N O

2 = 9.0132 = 9.01 g N2O

3-16

H2O:

( ) 3 2 23 23 2 3 2 2

1 mol Ca(NO ) 4 mol H O 18.02 g H O16.8 g Ca(NO )

164.10 g Ca(NO ) 1 mol Ca(NO ) 1 mol H O

2 = 7.3793 = 7.38 g H2O 3.56 Plan: Multiply the yield of the first step by that of the second step to get the overall yield. Solution: It is simpler to use the decimal equivalents of the percent yields, and then convert to percent using 100%. (0.73) (0.68) (100%) = 49.64 = 50.% 3.57 It is simpler to use the decimal equivalents of the percent yields, and then convert to percent using 100%. (0.48) (0.73) (100%) = 35.04 = 35% 3.58 Plan: Balance the chemical equation using the formulas of the substances. Determine the yield (theoretical yield)

for the reaction from the mass of tungsten(VI) oxide. Use the density of water to determine the actual yield of water in grams. The actual yield divided by the yield just calculated (with the result multiplied by 100%) gives the percent yield. Solution: (Rounding to the correct number of significant figures will be postponed until the final result.)

The balanced chemical equation is: WO3(s) + 3 H2(g) W(s) + 3 H2O(l) Theoretical yield of H2O:

( ) 3 233 3

1 mol WO 3 mol H O 18.02 g H O45.5 g WO

231.9 g WO 1 mol WO 1 mol H O

2

2 = 10.60686 g H2O

Actual yield, in grams, of H2O:

( ) 222

1.00 g H O9.60 mL H O

1 mL H O

= 9.60 g H2O

Calculate the percent yield:

ActualYield x 100%Theoretical Yield

= 2

2

9.60 g H Ox 100%

10.60686 g H O

= 90.5075 = 90.5%

3.59 Balance the chemical equation using the formulas of the substances. Determine the yield (theoretical yield) for the

reaction from the mass of phosphorus trichloride. The actual yield divided by the yield just calculated (with the result multiplied by 100%) gives the percent yield.

(Rounding to the correct number of significant figures will be postponed until the final result.) The balanced chemical equation is: PCl3(l) + 3 H2O(l) H3PO3(aq) + 3 HCl(g) Theoretical yield of H2O:

( ) 333 3

1 mol PCl 3 mol HCl 36.46 g HCl200. g PCl137.32 g PCl 1 mol PCl 1 mol HCl

= 159.3067 g HCl (Theoretical yield)

The actual yield is given (128 g HCl). Calculate the percent yield:

ActualYield x 100%Theoretical Yield

= 128 g HCl x 100%

159.3067 g HCl

= 80.3481586 = 80.3%

3.60 Plan: Write the balanced chemical equation. Since quantities of reactants are present, we must determine which is limiting. Only 80.0% of the calculated amounts of products will form. (Rounding to the correct number of significant figures will be postponed until the final result.) Solution: CH4(g) + Cl2(g) CH3Cl(g) + HCl(g) Mole CH3Cl from CH4 = ( ) 344

4 4

1 mol CH Cl1 mol CH20.5 g CH

16.04 g CH 1 mol CH

= 1.278055 mol CH3Cl

3-17

Mole CH3Cl from Cl2 = ( ) 3222 2

1 mol CH Cl1 mol Cl45.0 g Cl

70.90 g Cl 1 mol Cl

= 0.634697 mol CH3Cl

Chlorine is the limiting reactant.

Grams CH3Cl = ( ) 3 3222 2 3

1 mol CH Cl 50.48 g CH Cl1 mol Cl 75.0%45.0 g Cl70.90 g Cl 1 mol Cl 1 mol CH Cl 100%

= 24.02962 = 24.0 g CH3Cl The beginning of the calculation is repeated to emphasize that the second part of the problem is simply an extension of the first part. There is no need to repeat the entire calculation as only the final step(s) times the answer of the first part will give the final answer to this part. 3.61 First, balance the chemical equation: 3 Ca(s) + N2(g) Ca3N2(s) Mole Ca3N2 from Ca = ( ) 3 21 mol Ca N1 mol Ca56.6 g Ca 40.08 g Ca 3 mol Ca

= 0.470725 mol Ca3N2

Mole Ca3N2 from N2 = ( ) 3 2222 2

1 mol Ca N1 mol N30.5 g N

28.02 g N 1 mol N

= 1.0885 mol CH3Cl

Ca is the limiting reactant.

Grams Ca3N2 = ( ) 3 2 3 23 2

1 mol Ca N 148.26 g Ca N1 mol Ca 93.0%56.6 g Ca40.08 g Ca 3 mol Ca 1 mol Ca N 100%

= 64.90444 = 64.9 g Ca3N2 3.62 Plan: The first step is to determine the chemical formulas so a balanced chemical equation can be written. The limiting reactant must be determined. Finally, the mass of CF4 is determined from the limiting reactant. Solution: (Rounding to the correct number of significant figures will be postponed until the final result.) The balanced chemical equation is: (CN)2(g) + 7 F2(g) 2 CF4(g) + 2 NF3(g) Mole CF4 from (CN)2 = ( ) 22

2 2

1 mol (CN) 2 mol CF60.0 g (CN)

52.04 g (CN) 1 mol (CN)

4 = 2.30592 mol CF4

Mole CF4 from F2 = ( ) 222 2

1 mol F 2 mol CF60.0 g F

38.00 g F 7 mol F

4 = 0.4511278 mol CF4

F2 is the limiting reactant, and will be used to calculate the yield.

Grams CF4 = ( ) 2 422 2

1 mol F 2 mol CF 88.01 g CF60.0 g F

38.00 g F 7 mol F 1 mol CF

4

4

= 39.70376 = 39.7 g CF4

3.63 a) The reaction is 2 Cl2O(g) 2 Cl2(g) + O2(g).

Both oxygen and chlorine are diatomic and the ratio between Cl2 and O2 is 2:1. Scene A best represents the product mixture as there are three O2 molecules and twice as many, or six Cl2 molecules in Scene A. This is the proper mole ratio. Scene B shows oxygen and chlorine atoms and Scene C shows oxygen atoms. Oxygen and chlorine atoms are NOT products of this reaction.

b) 2 Cl2O(g) 2 Cl2(g) + O2(g) c)

( ) 232 22 2

1 mol O 2 mol Cl O 6.02 x 10 molecules Cl O0.050 mol oxygen atoms6 oxygen atoms1 oxygen atom 2 mol oxygen atoms 1 mol O 1 mol Cl O

2

= 1.806 x 1023 = 1.8 x 1023 Cl2O molecules

3-18

3.64 Determine the balanced chemical equation: 2 C4H10(g) + 13 O2(g) 8 CO2(g) + 10 H2O(g) a) Grams O2 = ( ) 4 10 4 10 2 24 10

4 10 4 10 4 10 2

0.579 g C H 1 mol C H 13 mol O 32.00 g O5.50 mL C H

1 mLC H 58.12 g C H 2 mol C H 1 mol O

= 11.3967 = 11.4 g O2

b) Moles H2O = ( ) 4 10 4 10 24 104 10 4 10 4 10

0.579 g C H 1 mol C H 10 mol H O5.50 mL C H

1 mL C H 58.12 g C H 2 mol C H

= 0.27396 = 0.274 mol H2O

c) Total moles = ( ) 4 10 4 10 24 104 10 4 10 4 10

0.579 g C H 1 mol C H 8 mol CO5.50 mL C H

1 mL C H 58.12 g C H 2 mol C H

+

( ) 4 10 4 10 24 104 10 4 10 4 10

0.579 g C H 1 mo l C H 10 mol H O5.50 mL C H

1 mL C H 58.12 g C H 2 mol C H

= 0.49312629 mol

Total molecules = ( ) 236.022 x 10 molecules0.49312629 mol1 mol

= 2.96961x 1023 = 2.97 x 1023 molecules 3.65 Balancing the chemical equation gives: 2 NaH(s) + B2H6(g) 2 NaBH4(s) Determine the limiting reactant. (Rounding to the correct number of significant figures will be postponed until the final result.)

Mole NaBH4 from NaH = ( ) 42 mol NaBH1 mol NaH7.98 g NaH 24.00 g NaH 2 mol NaH

= 0.3325 mol NaBH4

Mole NaBH4 from B2H6 = ( ) 2 6 42 62 6 2 6

1 mol B H 2 mol NaBH8.16 g B H

27.67 g B H 1 mol B H

= 0.58981 mol NaBH4

NaH is the limiting reactant, and will be used to calculate the yield.

( ) 4 44

2 mol NaBH 37.83 g NaBH1 mol NaH 88.5%7.98 g NaH24.00 g NaH 2 mol NaH 1 mol NaBH 100%

= 11.13195 = 11.1 g NaBH4

3.66 a) Solution B has the highest molarity as it has the largest number of particles, 12, in a volume of 50 mL.

b) Solutions A and F both have 8 particles in a volume of 50 mL and thus the same molarity. Solutions C, D, and E all have 4 particles in a volume of 50 mL and thus have the same molarity. c) Mixing Solutions A and C results in 12 particles in a volume of 100 mL. That is a lower molarity than that of Solution B which has 12 particles in a volume of 50 mL or 24 particles in a volume of 100 mL. d) Adding 50 mL to Solution D would result in 4 particles in a total volume of 100 mL; adding 75 mL to Solution F would result in 4 particles in a volume of 100 mL. The molarity of each solution would be the same. e) Solution A has 8 particles in a volume of 50 mL while Solution E has the equivalent of 4 particles in a volume of 50 mL. The molarity of Solution E is half that of Solution A. Therefore half of the volume, 12.5 mL, of Solution E must be evaporated. When 12.5 mL of solvent is evaporated from Solution E, the result will be 2 particles in 12.5 mL or 8 particles in 50 mL as in Solution A.

3.67 Plan: The spheres represent particles of solute and the amount of solute per given volume of solution determines its concentration. Molarity = moles of solute/volume (L) of solution. Solution: a) Box C has more solute added because it contains 2 more spheres than Box A contains. b) Box B has more solvent because solvent molecules have displaced two solute molecules. c) Box C has a higher molarity, because it has more moles of solute per volume of solution. d) Box B has a lower concentration (and molarity), because it has fewer moles of solute per volume of solution.

3-19

3.68 Plan: In all cases, the definition of molarity (solution of L

solute moles M = ) will be important. Volume must be expressed in liters. The molar mass is used to convert moles to grams. The chemical formulas must be written to determine the molar mass.

Solution:

a) Grams Ca(C2H3O2)2 = ( ) 3 2 3 2 2 2 3 2 22 3 2 2

0.267 mol Ca(C H O ) 158.17 g Ca(C H O )10 L185.8 mL1 mL 1 L 1 mol Ca(C H O )

= 7.84659 = 7.85 g Ca(C2H3O2)2

b) Moles KI = ( ) 1 mol KI21.1 g KI166.0 g KI

= 0.127108 moles KI

Volume = ( ) 3 10 L500. mL1 mL

= 0.500 L

Molarity KI = L 0.500

KI mol 127108.0 = 0.254216 = 0.254 M KI

c) Moles NaCN = (0.850 mol NaCN 145.6 L1 L

) = 123.76 = 124 mol NaCN 3.69 Molarity is a convenient way of expressing concentration; and, it is useful to use the definition (mole/L) in place of M.

a) Milliliters KOH solution = ( ) 31 mol KOH 1 L 1 mL8.42 g KOH 56.11 g KOH 2.26 M 10 L

= 66.39928 = 66.4 mL KOH solution

b) Number Cu2+ ions = ( )2 232 22

2.3 mol CuCl 1 mol Cu 6.022 x 10 Cu ions52 LL 1 mol CuCl 1 mol Cu

+ ++

2

= 7.2023 x 1025 = 7.2 x 1025 Cu2+ ions

c) M glucose = 135 mmol glucose275 mL

= 0.490909 = 0.491 M glucose

3.70 Plan: These are dilution problems. Dilution problems can be solved by converting to moles and using the new volume; however, it is much easier to use M1V1 = M2V2. Part (c) may be done as two dilution problems or as a mole problem. The dilution equation does not require a volume in liters; it only requires that the volume units match.

Solution: a) M1 = 0.250 M KCl V1 = 37.00 mL M2 = ? V2 = 150.00 mL M1V1 = M2V2 (0.250 M)(37.00 mL) = (M2)(150.0 mL)

(0.250 )(37.00 mL)150.0 mLM = M2 M2 = 0.061667 = 0.0617 M KCl

b) M1 = 0.0706 M (NH4)2SO4 V1 = 25.71 mL M2 = ? V2 = 500.00 mL M1V1 = M2V2 (0.0706 M)(25.71 mL) = (M2)(1500.0 mL)

(0.0706 )(25.71 mL)500.0 mL

M = M2 M2 = 0.003630 = 0.00363 M (NH4)2SO4

c) When working this as a mole problem it is necessary to find the individual number of moles of sodium ions in each separate solution. (Rounding to the proper number of significant figures will only be done for the final answer.)

Moles Na+ from NaCl solution = ( ) 310 L 0.348 mol NaCl 1 mol Na3.58 mL1 mL 1 L 1 mol NaCl

+

= 0.00124584 mol Na+

3-20

Moles Na+ from Na2SO4 solution = ( ) 23 2 42 4

6.81 x 10 mol Na SO10 L 2 mol Na500. mL1 mL 1 L 1 mol Na SO

+

= 0.0681 mol Na+

Molarity of Na+ = ( )

( ) 30.00124584 0.0681 mol Na 1 mL

3.58 500. mL 10 L

+

+ + = 0.1377058 = 0.138 M Na

+ ions

3.71 These are dilution problems using the equation M1V1 = M2V2. a) M1 = 2.050 M Cu(NO3)2 V1 = ? M2 = 0.8543 M Cu(NO3)2 V2 = 750.0 mL V1 = M2V2 / M1 = (0.8543 M) (750.0 mL) / (2.050 M) = 312.5488 = 312.5 mL

b) 1.63 M CaCl2 gives: M Cl = 22

1.63 mol CaCl 2 mol Cl1 L 1 mol CaCl

= 3.26 M Cl ions

M1 = 3.26 M Cl V1 = ? M2 = 2.86 x 102 M Cl ions V2 = 350. mL V1 = M2V2 / M1 = (2.86 x 102 M) (350. mL) / (3.26 M) = 3.07055= 3.07 mL

c) M1 = 0.155 M Li2CO3 V1 = 18.0 mL M2 = 0.0700 M Li2CO3 V2 = ? V2 = M1V1 / M2 = (0.155 M) (18.0 mL) / (0.0700 M) = 39.8571 = 39.9 mL 3.72 Plan: Use the density of the solution to find the mass of 1 L of solution. The 70.0% by mass translates to

70.0 g solute/100 g solution and is used to find the mass of HNO3 in 1 L of solution. Convert mass of HNO3 to moles to obtain moles/L, molarity.

Solution:

a) Mass HNO3 per liter = 3370.0 g HNO1.41 g Solution 1 mL

1 mL 100 g Solution10 L

= 987 g HNO3/L

b) Molarity of HNO3 = 3 333

70.0 g HNO 1 mol HNO1.41 g Solution 1 mL1 mL 100 g Solution 63.02 g HNO10 L

= 15.6617 = 15.7 M HNO3

3.73 a) Moles per milliliter = 3

2 418.3 mol H SO 10 L1 L 1 mL

= 1.83 x 102 mol H2SO4/mL

b) To calculate the mass percent, the mass of H2SO4 in each milliliter (1.84 g) of solution is needed.

Mass percent = ( )32 4 2 42 4

18.3 mol H SO 98.09 g H SO10 L 1 mL100%1 L 1 mL 1 mol H SO 1.84 g solution

= 97.5569 = 97.6% H2SO4 by mass 3.74 Plan: Convert the mass of calcium carbonate to moles, and use the balanced chemical equation to find the moles

of hydrochloric acid required. The moles of acid along with the molarity of the acid will give the volume required. The molarity of the solution is given in the calculation as mol/L. Solution:

2 HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l) Volume required = ( ) 33 3

3 3

1 mol CaCO 2 mol HCl 1 L 1 mL16.2 g CaCO100.09 g CaCO 1 mol CaCO 0.383 mol HCl 10 L

= 845.1923 = 845 mL HCl solution

3-21

3.75 Convert the molarity of sodium hydroxide to moles, and use the balanced chemical equation to find the moles of sodium dihydrogen phosphate required. The moles of sodium dihydrogen phosphate, along with its molar mass, will give the grams required. The molarity of the solution is given in the calculation as mol/L. NaH2PO4(s) + 2 NaOH(aq) Na3PO4(aq) + 2 H2O(l) Mass of sodium dihydrogen phosphate =

( ) 3 2 4 2 42 4

1 mol NaH PO 119.98 g NaH PO10 L 0.285 mol NaOH43.74 mL1 mL 1 L 2 mol NaOH 1 mol NaH PO

= 0.747829 = 0.748 g NaH2PO4 3.76 Plan: The first step is to write and balance the chemical equation for the reaction. Use the molarity and volume of each of the reactants to determine the moles of each as a prelude to determining which is the limiting reactant. Use the limiting reactant to determine the mass of barium sulfate that will form. Solution: The balanced chemical equation is: BaCl2(aq) + Na2SO4(aq) BaSO4(s) + 2 NaCl(aq) The mole and limiting reactant calculations are:

Moles BaSO4 from BaCl2 = ( ) 3 22

0.160 mol BaCl 1 mol BaSO10 L35.0 mL1 mL 1 L 1 mol BaCl

4 = 0.00560 mol BaSO4

Moles BaSO4 from Na2SO4 = ( ) 3 2 4 42 4

0.065 mol Na SO 1 mol BaSO10 L58.0 mL1 mL 1 L 1 mol Na SO

= 0.00377 mol BaSO4

Sodium sulfate is the limiting reactant.

Grams BaSO4 = ( ) 3 2 4 4 42 4 4

0.065 mol Na SO 1 mol BaSO 233.4 g BaSO10 L58.0 mL1 mL 1 L 1 mol Na SO 1 mol BaSO

= 0.879918 = 0.88 g BaSO4 3.77 The first step is to write and balance the chemical equation for the reaction. Use the molarity and volume of each of the reactants to determine the moles of each as a prelude to determining which is the limiting reactant. Use the limiting reactant to determine the mass of the other substance consumed. H2SO4(aq) + 2 NaOH(aq) Na2SO4(aq) + 2 H2O(l) We can use either product to determine the limiting reactant. We will use sodium sulfate.

Moles Na2SO4 from H2SO4 = ( ) 3 2 4 2 42 4

0.210 mol H SO 1 mol Na SO10 L350.0 mL1 mL 1 L 1 mol H SO

= 0.0735 mol Na2SO4

Moles Na2SO4 from NaOH = ( ) 2 41 mol Na SO0.196 mol NaOH0.500 L 1 L 2 mol NaOH

= 0.0490 mol Na2SO4

NaOH is the limiting reactant. Finish the problem using the limiting reactant. Moles H2SO4 remaining = Initial moles Moles reacting with NaOH

= ( ) 3 2 40.210 mol H SO10 L350.0 mL1 mL 1 L

( ) 2 41 mol H SO0.196 mol NaOH0.500 L

1 L 2 mol NaOH

= 0.0245 mol H2SO4 3.78 Plan: The first part of the problem is a simple dilution problem (M1V1 = M2V2). The second part requires the

molar mass of the HCl along with the molarity. Solution: a) M1 = 11.7 M V1 = ? M2 = 3.5 M V2 = 3.0 gal M1V1 = M2V2 (11.7 M)(V1) = (3.5 M)(3.0 gal)

(3.5 )(3.0 gal)11.7 M

M = V1 V1 = 0.897436 gallons (unrounded)

3-22

Instructions: Be sure to wear goggles to protect your eyes! Pour approximately 2.0 gallons of water into the container. Add slowly and with mixing 0.90 gallons of 11.7 M HCl into the water. Dilute to 3.0 gallons with water.

b) Volume needed = ( ) 31 mol HCl 1 L 1 mL9.66 g HCl 36.46 g HCl 11.7 mol HCl 10 L

= 22.64512 = 22.6 mL muriatic acid solution 3.79 Plan: The moles of narceine and the moles of water are required. We can assume any mass of narceine hydrate (we will use 100 g), and use this mass to determine the mass of water present and convert the mass to moles of the hydrate. The mass of water will be converted to moles. Finally, the ratio of the moles of hydrate to moles of water will give the amount of water present.

Solution:

Moles narceine hydrate = ( ) 1 mol narceine hydrate100 g narceine hydrate499.52 g narceine hydrate

= 0.20019 mol narceine hydrate

Moles H2O = ( ) 22

10.8% H O 1 mol H O100 g narceine hydrate

100% narceine hydrate 18.02 g H O

2 = 0.59933 mol H2O The ratio of water to hydrate is: (0.59933 mol)/(0.20019 mol) = 3 Thus, there are three water molecules per mole of hydrate. The formula for narceine hydrate is narceine3H2O. 3.80 Plan: Determine the formula, then the molar mass of each compound. Determine the mass of hydrogen in each formula. The mass of hydrogen divided by the molar mass of the compound (with the result multiplied by 100%) will give the mass percent hydrogen. Ranking, based on the percents, is easy. Solution: Name Chemical Molar mass Mass percent H formula (g/mol) [(mass H)/(molar mass)] x 100% Ethane C2H6 30.07 [(6 x 1.008)/(30.07)] x 100% = 20.11% H Propane C3H8 44.09 [(8 x 1.008)/(44.09)] x 100% = 18.29% H Cetyl palmitate C32H64O2 480.83 [(64 x 1.008)/(480.83)] x 100% = 13.42% H Ethanol C2H5OH 46.07 [(6 x 1.008)/(46.07)] x 100% = 13.13% H Benzene C6H6 78.11 [(6 x 1.008)/(78.11)] x 100% = 7.743% H The hydrogen percentage decreases in the following order: Ethane > Propane > Cetyl palmitate > Ethanol > Benzene

3.81 a) 2 H2S(g) + 3 O2(g) 2 SO2(g) + 2 H2O(g) b) 4 KClO3(s) KCl(s) + 3 KClO4(s) c) 3 H2(g) + Fe2O3(s) 2 Fe(s) + 3 H2O(g) d) 2 C2H6(g) + 7 O2(g) 4 CO2(g) + 6 H2O(g) e) 2 FeCl2(s) + 2 ClF3(g) 2 FeF3(s) + 3 Cl2(g) 3.82 Isobutylene + O2 CO2 + H2O Mole C = ( ) 22

2 2

1 mol CO 1 mol C2.657 g CO44.01 g CO 1 mol CO

= 0.06037 mol C

Mole H = ( ) 222 2

1 mol H O 2 mol H1.089 g H O18.02 g H O 1 mol H O

= 0.1209 mol H

Divide each mole value by the smaller value (mole C). C: 0.06037 mol / 0.06037 mol = 1.00 H: 0.1209 mol / 0.06037 mol = 2.00 This gives an empirical formula of CH2.

3-23

3.83 C7H8 + 9 O2(g) 7 CO2(g) + 4 H2O(g) Moles C7H8 = ( ) 7 8 7 87 8

7 8 7 8

0.867 g C H 1 mol C H20.0 mL C H

1 mL C H 92.13 g C H

= 0.1882123 mol C7H8 (unrounded)

a) Grams oxygen = ( ) 27 87 8 2

9 mol O 32.00 g O0.1882123 mol C H

1 mol C H 1 mol O

2 = 54.20514 = 54.2 g O2

b) Total moles of gas = ( )7 87 8

11 mol product gas0.1882123 mol C H1 mol C H

= 2.07034 = 2.07 mol of gas

The 11 mol of gas is an exact, not measured, number, so it does not affect the significant figures.

c) Molecules of water = ( ) 232 27 87 8 2

4 mol H O 6.022 x 10 H O molecules0.1882123 mol C H

1 mol C H 1 mol H O

= 4.53366 x 1023 = 4.53 x 1023 molecules H2O 3.84 Plan: If 100.0 g of dinitrogen tetroxide reacts with 100.0 g of hydrazine (N2H4), what is the theoretical yield of nitrogen if no side reaction takes place? First, we need to identify the limiting reactant. The limiting reactant can be used to calculate the theoretical yield. Determine the amount of limiting reactant required to produce 10.0 grams of NO. Reduce the amount of limiting reactant by the amount used to produce NO. The reduced amount of limiting reactant is then used to calculate an actual yield. The actual and theoretical yields will give the maximum percent yield.

Solution: Determining the limiting reactant:

N2 from N2O4 = ( ) 2 4 22 42 4 2 4

1 mol N O 3 mol N100.0 g N O

92.02 g N O 1 mol N O

= 3.26016 mol N2

N2 from N2H4 = ( )

42

2

42

4242 HN mol 2

N mol 3HN g 32.05

HN mol 1HN g 0.100 = 4.68019 mol N2

N2O4 is the limiting reactant.

Theoretical yield of N2 = ( ) 2 4 2 22 42 4 2 4 2

1 mol N O 3 mol N 28.02 g N100.0 g N O

92.02 g N O 1 mol N O 1 mol N

= 91.3497 g N2 (unrounded)

How much limiting reactant is used to produce 100.0 g NO?

Grams N2O4 used = ( ) 2 4 2 42 4

2 mol N O 92.02 g N O1 mol NO10.0 g NO30.01 g NO 6 mol NO 1 mol N O

= 10.221 g N2O4 (unrounded)

Determine the actual yield. Amount of N2O4 available to produce N2 = 100.0 g N2O4 mass of N2O4 required to produce 10.0 g NO 100.0 g 10.221 g = 89.779 g N2O4 (unrounded)

Actual yield of N2 = ( )

2

2

42

2

42

4242 N mol 1

N g 02.28ON mol 1

N mol 3ON g 92.02

ON mol 1ON g 779.89

= 82.01285 g N2 (unrounded) Theoretical yield = [Actual yield / theoretical yield] x 100% [(82.01285 g N2) / (91.3497 g N2)] x 100% = 89.7790 = 89.8% 3.85 Plan: Count the number of each type of molecule in the reactant box and in the product box. Subtract any molecules of excess reagent (molecules appearing in both boxes). The remaining material is the overall equation. This will need to be simplified if there is a common factor among the substances in the equation. The balanced chemical equation is necessary for the remainder of the problem.

3-24

Solution: a) The contents of the boxes give: 6 AB2 + 5 B2 6 AB3 + 2 B2 B2 is in excess, so two molecules need to be removed from each side. This gives: 6 AB2 + 3 B2 6 AB3 Three is a common factor among the coefficients, and all coefficients need to be divided by this value to give the final balanced equation: 2 AB2 + B2 2 AB3 b) B2 was in excess, thus AB2 is the limiting reactant.

c) Moles of AB3 from AB2 = ( )

2

32 AB mol 2

AB mol 2AB mol 0.5 = 5.0 mol AB3

Moles of AB3 from B2 = ( )

2

32 B mol 1

AB mol 2B mol 0.3 = 6.0 mol AB3

AB2 is the limiting reagent and maximum is 5.0 mol AB3.

d) Moles of B2 that react with 5.0 mol AB2 = ( )

2

22 AB mol 2

B mol 1AB mol 0.5 = 2.5 mol B2

The unreacted B2 is 3.0 mol 2.5 mol = 0.5 mol B2.

3.86 a) ( ) 85% NaCl4.0% ions100% ions

= 3.4% NaCl

b) % Na+ ions = ( ) 22.99 g Na3.4% NaCl58.44 g NaCl

+ = 1.3375 = 1.3% Na+ ions

% Cl ions = ( ) 35.45 g Cl3.4% NaCl58.44 g NaCl

= 2.062 = 2.1% Cl ions

c) M NaCl = 33.4 g NaCl 1.025 g Seawater 1 mL 1 mol NaCl

100 g Seawater 1 mL 58.44 g NaCl10 L

= 0.596338 = 0.60 M NaCl

3.87 a) False, a mole of one substance has the same number of units as a mole of any other substance. b) True c) False, a limiting-reactant problem is presented when the quantity of available material is given for more than one reactant. d) True 3.88 Plan: Count the total number of spheres in each box. The number in box A divided by the volume change in each part will give the number we are looking for and allow us to match boxes. Solution: The number in each box is: A = 12, B = 6, C = 4, and D = 3. a) When the volume is tripled, there should be 12/3 = 4 spheres in a box. This is box C. b) When the volume is doubled, there should be 12/2 = 6 spheres in a box. This is box B. c) When the volume is quadrupled, there should be 12/4 = 3 spheres in a box. This is box D. 3.89 a) Equal Equal numbers of particles b) O3 0.4 mol O3 has a greater mass c) SO2 (4.0 g N2O4) (1 mol/92.02 g N2O4) = 0.043 mol N2O4 (3.3 g SO2) (1 mol/64.07 g SO2) = 0.052 mol SO2 d) F2 (0.6 mol C2H4) (28.05 g/1 mol C2H4) = 17 g (0.6 mol F2) (38.00 g/mol F2) = 23 g Note that if each of these values is properly rounded to one significant figure, the answers are identical.

3-25

e) MgCl2 (2.2 mol MgCl2) (3 ions/mol MgCl2) = 6.6 mol (2.3 mol NaClO3) (2 ions/mol NaClO3) = 4.6 mol f) H2O The compound with the lower molar mass will have more molecules in a given mass. g) Equal (0.500 mol NaBr/L) (0.500 L) (1 mol Na+/mol NaBr) = 0.250 mol Na+ (0.0146 kg NaCl) (103 g/1 kg) (1 mol NaCl/58.44 g) (1 mol Na+/mol NaCl) = 0.250 mol Na+ Multiply the moles by Avogadros number to find the number of ions. This step is unnecessary since the moles are equal. h) 238U The heavier atoms will give a greater total mass. 3.90 P4S3(s) + 8 O2(g) P4O10(s) + 3 SO2(g) a) 1 molecule of P4S3 reacts with 8 molecules of O2 to produce 1 molecule of P4O10 and 3 molecules of SO2. b) 1 mol P4S3 reacts with 8 mol O2 to produce 1 mol P4O10 and 3 mol SO2. c) 220.09 g of P4S3 react with 256.00 g of O2 to produce 283.88 g of P4O10 and 192.21 g of SO2. 3.91 2 H2(g) + O2(g) 2 H2O(g) Mass H2 = ( ) 3 2 2 22

2 2 2

1 mol H O 2 mol H 2.016 g H10 g 100%105 kg H O1 kg 18.02 g H O 2 mol H O 1 mol H 98.8%

= 1.18896 x 104 = 1.19 x 104 g H2 3.92 Plan: This problem may be done as two dilution problems with the two final molarities added, or, as done here, it may be done by calculating, then adding the moles and dividing by the total volume. Solution:

M KBr = Total Moles KBrTotal Volume

= Moles KBr from Solution 1 Moles KBr from Solution 2Volume Solution 1 Volume Solution 2

++

M KBr = ( ) ( )0.053 mol KBr 0.078 mol KBr0.200 L 0.550 L

1 L 1 L0.200 L 0.550 L

+ + = 0.071333 = 0.071 M KBr

3.93 a) Moles NH4Br = ( ) 444

1 mol NH Br0.588 g NH Br

97.94 g NH Br

= 0.0060037 = 0.00600 mol NH4Br

b) Number of K+ ions = ( ) 23333 3

1 mol KNO 1 mol K 6.022 x 10 K ions88.5 g KNO101.11 g KNO 1 mol KNO 1 mol K

+ ++

= 5.27096 x 1023 = 5.27 x 1023 K+ ions

c) Mass C3H8O3 = ( ) 3 8 33 8 33 8 3

92.09 g C H O5.85 mol C H O

1 mol C H O

= 538.7265 = 539 g C3H8O3

d) Volume CHCl3 = ( ) 333 3

119.37 g CHCl mL2.85 mol CHCl1 mol CHCl 1.48 g CHCl

= 229.868 = 230. mL CHCl3

e) Number of Na+ = ( ) 232 32 3

2 mol Na 6.022 x 10 Na ions2.11 mol Na CO1 mol Na CO 1 mol Na

+ ++

= 2.54128 x 1024 = 2.54 x 1024 Na+ ions

f) Number of Cd atoms = ( ) 6 210 g 1 mol Cd 6.022 x 10 Cd atoms25.0 g Cd1 g 112.4 g Cd 1 mol Cd

3

= 5.35765 x 1016 = 5.36 x 1016 Cd atoms

g) Number of F atoms = ( ) 2322

2 mol F 6.022 x 10 F atoms0.0015 mol F1 mol F 1 mol F

= 1.8066 x 1021 = 1.8 x 1021 F atoms

3-26

3.94 Neither A nor B has any XY3 molecules. Both C and D have XY3 molecules. D forms both XY3 and XY molecules. Only C has a single XY3 product, thus the answer is C. 3.95 Plan: Deal with the methane and propane separately, and combine the results. Balanced equations are needed for

each hydrocarbon. The total mass and the percentages will give the mass of each hydrocarbon. The mass of each hydrocarbon is changed to moles, and through the balanced chemical equation the amount of CO2 produced by each gas may be found. Summing the amounts of CO2 gives the total from the mixture. For Part b), let x and

252 x represent the masses of CH4 and C3H8, respectively. Solution:

a) The balanced chemical equations are: Methane: CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l) Propane: C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(l) Mass of CO2 from each:

Methane: ( ) 4 24 4

1 mol CH 1 mol CO 44.01 g CO25.0%200. g mixture100% 16.04 g CH 1 mol CH 1 mol CO

2

2

= 137.188 g CO2

Propane: ( ) 3 8 23 8 3 8 2

1 mol C H 3 mol CO 44.01 g CO75.0%200. g mixture100% 44.09 g C H 1 mol C H 1 mol CO

2 = 449.183 g CO2

Total CO2 = 137.188 g + 449.183 g = 586.371 = 586 g CO2 b) Since the mass of CH4 + the mass of C3H8 = 252 g, let x = mass of CH4 in the mixture and 252 x = mass of C3H8 in the mixture. Use mole ratios to calculate the amount of CO2 formed from x amount of CH4 and the amount of CO2 formed from 252 x amount of C3H8. The total mass of CO2 produced = 748 g.

The total moles of CO2 produced = ( ) 222

1 mol CO748 g CO

44.01 g CO

= 16.996 mol CO2

16.996 mol CO2 = ( ) ( ) 3 84 24 3 84 4 3 8

1 mol C H1 mol CH 1 mol CO 3 mol COx g CH + 252 x g C H

16.04 g CH 1 mol CH 44.09 g C H 1 mol C H

2

3 8

16.996 mol CO2 = 2 2x 3(252 x) mol CO + mol CO

16.04 44.09

16.996 mol CO2 = 2 2x 756 3x mol CO + mol CO

16.04 44.09

16.996 mol CO2 = 0.06234x mol CO2 + (17.147 0.06804x mol CO2) 16.996 = 17.147 0.0057x x = 26.49 g CH4 252 x = 252 g 26.49 g = 225.51 g C3H8

mass % CH4 = 426.49 g CH

100252 g mixture

= 10.5% CH4

mass % C3H8 = 3 8225.51 g C H

100252 g mixture

= 89.5% C3H8

3.96 Plan: If we assume a 100-gram sample of fertilizer, then the 30:10:10 percentages become the masses, in grams, of N, P2O5, and K2O. These masses may be changed to moles of substance, and then to moles of each element. To get the desired x:y:1.0 ratio, divide the moles of each element by the moles of potassium. Solution: A 100-gram sample of 30:10:10 fertilizer contains 30 g N, 10 g P2O5, and 10 g K2O.

Moles N = ( ) 1 mol N30g N14.01 g N

= 2.1413 mol N (unrounded)

3-27

Moles P = ( ) 2 52 52 5 2 5

1 mol P O 2 mol P10 g P O141.94 g P O 1 mol P O

= 0.14090 mol P (unrounded)

Moles K = ( ) 222 2

1 mol K O 2 mol K10 g K O94.20 g K O 1 mol K O

= 0.21231 mol K (unrounded)

This gives a ratio of 2.1413:0.14090:0.21231 The ratio must be divided by the moles of K and rounded. (2.1413/0.21231):(0.14090/0.21231):(0.21231/0.21231) 10.086:0.66365:1.000 or 10:0.66:1.0 3.97 Assume a 100 g sample. 10:10:10 indicates 10 g N, 10 g P2O5 and 10 g K2O.

Moles N = ( ) 1 mol N10g N14.01 g N

= 0.713776 mol N

Moles P = ( ) 2 52 52 5 2 5

1 mol P O 2 mol P10 g P O141.94 g P O 1 mol P O

= 0.14090 mol P

Moles K = ( ) 222 2

1 mol K O 2 mol K10 g K O94.20 g K O 1 mol K O

= 0.21231 mol K

To obtain 0.713776 mol N from (NH4)2SO4:

( ) 4 2 4 4 2 44 2 4

1 mol (NH ) SO 132.15 g (NH ) SO0.713776 mol N

2 mol N 1 mol (NH ) SO

= 47.1627 g (NH4)2SO4

mass % (NH4)2SO4 = 4 2 447.1627 g (NH ) SO

100100 g mixture

= 47.1627% = 47.2% (NH4)2SO4

To obtain 0.14090 mol P from (NH4)2HPO4:

( ) 4 2 4 4 2 44 2 4

1 mol (NH ) HP O 132.06 g (NH ) HP O0.14090 mol P

1 mol P 1 mol (NH ) HP O

= 18.6073 g (NH4)2HPO4

mass % (NH4)2HPO4 = 4 2 418.6073 g (NH ) HPO

100100 g mixture

= 18.6073% = 18.6% (NH4)2HPO4

To obtain 0.21231 mol K from KCl:

( ) 1 mol KCl 74.55 g KCl0.21231 mol K1 mol K 1 mol KCl

= 15.8277 g KCl

mass % KCl = 15.8277 g KCl 100100 g mixture

= 15.8277% = 15.8% KCl

3.98 SrX2(aq) + H2SO4(aq) SrSO4(s) + 2 HX(aq) 0.652 g 0.755 g

Moles SrX2 = ( ) 444 4

1 mol SrSO 1 mol SrX0.755 g SrSO

183.69 g SrSO 1 mol SrSO

2 = 0.004110186 mol SrX2 (unrounded)

The 0.652 g sample of SrX2 = 0.004110186 mol

SrX2 = mol 60.00411018g 652.0

= 158.630 g/mol = molar mass

Molar mass of X2 = 158.630 g molar mass of Sr Molar mass of X2 = 158.630 g 87.62 g = 71.01 g = X2 Molar mass of X = 71.01 g/2 = 35.505 = 35.5 g/mol = Cl The original halide formula is SrCl2.

3-28

3.99 Plan: Assume 100 grams of mixture. This means the mass of each compound, in grams, is the same as its percentage. Find the mass of C from CO and from CO2 and add these masses together. Solution: 100 g of mixture = 35 g CO and 65 g CO2.

C from CO = ( ) 1 mol CO 1 mol C 12.01 g C35.0 g CO28.01 g CO 1 mol CO 1 mol C

= 15.007 g C (unrounded)

C from CO2 = ( ) 222 2

1 mol CO 1 mol C 12.01 g C65.0 g CO44.01 g CO 1 mol CO 1 mol C

= 17.738 g C (unrounded)

Mass percent C = ( )15.007 17.738 g

x 100%100 g Sample

+ = 32.745 = 32.7% C

3.100 SiH4 + N2F4 SiF4 + N2 + 2H2 Moles of SiH4 = 3(1.25 x 102 mol) = 0.0375 mol SiH4 Moles of N2F4 = 3(1.25 x 102 mol) = 0.0375 mol N2F4

Mass of SiF4 = ( ) 4 444 4

1 mol SiF 104.09 g SiF 87%0.0375 mol SiH1 mol SiH 1 mol SiF 100%

= 3.3959 = 3.4 g SiF4 3.101 Ferrocene + ? O2(g) CO2 + H2O 0.9437 g 2.233 g 0.457 g

Moles C = ( ) 222 2

1 mol CO 1 mol C2.233 g CO44.01 g CO 1 mol CO

= 0.050738 mol C (unrounded)

Moles H = ( ) 222 2

1 mol H O 2 mol H0.457 g H O18.02 g H O 1 mol H O

= 0.050721 mol H (unrounded)

Mass of Fe = mass of sample mass of C mass of H = 0.9437 g Fe, C, H (0.050738 mol C) (12.01 g C/mol C) (0.050721 mol H) (1.008 g H/mol H) = (0.9437 0.60936 0.05113) g = 0.28321 g Fe (unrounded)

Moles Fe = (0.28321 g Fe) (1 mol Fe/55.85 g Fe) = 0.0050709 mol Fe Divide all moles by the smallest (Fe): C: (0.050738 mol/0.0050709 mol) = 10.0 H: (0.050721 mol/0.0050709 mol) = 10.0 Fe: (0.0050709 mol/0.0050709 mol) = 1.0 Thus, the empirical formula is: C10H10Fe 3.102 Plan: Determine the molecular formula from the figure. Once the molecular formula is known, use the

periodic table to determine the molar mass. Convert the volume in (b) from quarts to mL and use the density to convert from mL to mass in grams. Take 6.82% of that mass and use the molar mass to convert to moles.

Solution: a) The formula of citric acid obtained by counting the number of carbon atoms, oxygen atoms, and hydrogen atoms is C6H8O7. Molar mass = (6 x 12.01) + (8 x 1.008) + (7 x 16.00) = 192.12 g/mol

b) Determine the mass of citric acid in the lemon juice, and then use the molar mass to find the moles.

Moles C6H8O7 = ( ) 6 8 73 1 mol C H O1 L 1 mL 1.09 g 6.82%1.50 qt 1.057 qt mL 100% 192.12 g acid10 L

= 0.549104 = 0.549 mol C6H8O7

3-29

3.103 a) Moles of each element are needed. These may be determined from the mass of each element. The mass of Pt is given and the mass of F is (0.519 0.327) g F = 0.192 g F. Calculating the moles of each element gives: Moles Pt = (0.327 g Pt) (1 mol Pt /195.1 g Pt) = 0.001676 mol Pt (unrounded) Moles F = (0.192 g F) (1 mol F/19.00 g F) = 0.010105 mol F (unrounded) Dividing each by the smaller value (Pt) gives: Pt: (0.001676 mol)/(0.001676 mol) = 1.0 F: (0.010105 mol)/(0.001676 mol) = 6.0 Empirical formula = PtF6 b) Determine the moles of Xe and of PtF6. Moles PtF6 = (0.265 g PtF6) (1 mol PtF6/309.1 g PtF6) = 0.0008573 mol PtF6 (unrounded) Moles Xe = [(0.378 0.265)g Xe] (1 mol Xe/131.3 g Xe) = 0.0008606 mol Xe (unrounded) The moles are essentially identical so dividing by the smaller value gives 1 in each case. The empirical formula is: XePtF6. c) This problem can be solved as a system of two equations and two unknowns. The two equations are: The two unknowns are: Xe(g) + 2 F2(g) XeF4(s) x = mol XeF4 produced Xe(g) + 3 F2(g) XeF6(s) y = mol XeF6 produced Moles of Xe consumed = 1.85 x 104 mol present 9.00 x 106 mol excess = 1.76 x 104 mol Xe Then x + y = 1.76 x 104 mol Xe consumed 2 x + 3 y = 5.00 x 104 mol F2 consumed Solve for x using the first equation and substitute the value of x into the second equation: x = 1.76 x 104 y 2 (1.76 x 104 y) + 3 y = 5.00 x 104 (3.52 x 104) 2 y + 3 y = 5.00 x 104 y = (5.00 x 104) (3.52 x 104) = 1.48 x 104 mol XeF6 x = (1.76 x 104) (1.48 x 104) =2.8 x 105 mol XeF4 Convert moles of each product to grams using the molar masses (unrounded values): Mass XeF4 = (2.8 x 105 mol XeF4) x (207.3 g/mol) = 5.8044 x 103 g XeF4 Mass XeF6 = (1.48 x 104 mol XeF6) x (245.3 g/mol) = 3.63044 x 102 g XeF6 Calculate the percent of each compound using the total weight of the products: (5.8044 x 103 + 3.63044 x 102) g = 0.0421088 g % XeF4 = [(5.8044 x 103 g)/ 0.0421088 g)] x 100% = 13.784 = 14% XeF4 % XeF6 = [(3.63044 x 102 g)/(0.0421088 g)] x 100% = 86.2157 = 86.2% XeF6 3.104 Plan: Use the mass percent to find the mass of heme in the sample; use the molar mass to convert the mass of

heme to moles. Then find the mass of Fe in the sample. Solution:

a) Grams of heme = ( ) 6.0% heme0.65 g hemoglobin100% hemoglobin

= 0.039 g heme

b) Mole of heme = ( ) 1 mol heme0.039 g heme616.49 g heme

= 6.32614 x 105 = 6.3 x 105 mol heme

c) Grams of Fe = ( )5 1 mol Fe 55.85 g Fe6.32614 x 10 mol heme 1 mol heme 1 mol Fe = 3.5331 x 103 = 3.5 x 103 g Fe

d) Grams of hemin = ( )5 1 mol hemin 651.94 g hemin6.32614 x 10 mol heme 1 mol heme 1 mol hemin = 4.1243 x 102 = 4.1 x 102 g hemin

3.105 Mn:O ratio In sample: 1.00/ 1.42 or 0.704 In braunite: 2.00/ 3.00 or 0.667 In manganosite: 1.00/ 1.00 or 1.00

3-30

a) The total amount of ore is equal to the amount of braunite (B) + the amount of manganosite (M). B + M = 1.00 M = 1.00 B The amount of Mn is dependent on the samples composition. M(1.00) + B(0.667) = 0.704 (1.00 B) (1.00) + B(0.667) = 0.704 B = 0.888889 mol braunite M = 0.111111 mol manganosite

Mass of braunite = 0.888889 mol 157.88 g1 mol

= 140.338 g braunite

Mass of manganosite (g) = 0.111111 mol 70.94 g1 mol

)

= 7.88221 g manganosite

Mass percent braunite = [140.338 g / (140.338 g + 7.88221 g)] x 100% = 94.6821% Mass percent manganosite = [7.88221 g / (140.338 g + 7.88221 g)] x 100% = 5.3179% Mass of braunite = (542.3 g) (94.6821%/ 100%) = 513.461 = 514 g braunite Mass of manganosite = (542.3 g) (5.3179% / 100%) = 28.839 = 28.8 g manganosite

b) Mn3+/ Mn2+ =( )(

2 0.888890.11111

= 16.000 = 16.0

3.106 a) 5 Ca(OH)2(aq) + 3 H3PO4(aq) Ca5(PO4)3(OH)(s) + 9 H2O(l)

b) Find the limiting reagent.

Mole Ca5(PO4)3(OH) from Ca(OH)2 = ( ) 5 4 3222 2

1 mol Ca (PO ) (OH)1 mol Ca(OH)100. g Ca(OH)

74.10 g Ca(OH) 5 mol Ca(OH)

= 0.2699055 mol Ca5(PO4)3(OH) Mole Ca5(PO4)3(OH) from H3PO4 =

( ) 3 4 3 4 5 4 33 43 4 3 4 3 4

85 g H PO 1 mol H PO 1 mol Ca (PO ) (OH)100. g H PO solution

100. g H PO solution 97.99 g H PO 3 mol H PO

= 0.2891452 mol Ca5(PO4)3(OH) Ca(OH)2 is the limiting reactant, and will be used to calculate the yield.

( ) 5 4 3 5 4 3222 2 5 4

1 mol Ca (PO ) (OH) 502.32 g Ca (PO ) (OH)1 mol Ca(OH)100. g Ca(OH)

74.10 g Ca(OH) 5 mol Ca(OH) 1 molCa (PO ) (OH) 3

= 135.57893 = 140 g Ca5(PO4)3(OH)

3.107 a) Mole aspirin from C7H6O3 = ( ) 7 6 3 9 8 47 6 37 6 3 7 6 3

1 mol C H O 1 mol C H O3.077 g C H O

138.12 g C H O 1 mol C H O

= 0.0222777 mol C9H8O4 (unrounded)

Mole aspirin from C4H6O3 = ( ) 4 6 3 9 8 44 6 34 6 3 4 6 3

1 mol C H O 1 mol C H O1.080 g5.50 mL C H OmL 102.09 g C H O 1 mol C H O

= 0.058183955 mol C9H8O4 (unrounded) The limiting reactant is C7H6O3

b) First, calculate the theoretical yield from the limiting reagent:

Grams C9H8O4 = ( ) 7 6 3 9 8 4 9 8 47 6 37 6 3 7 6 3 9 8 4

1 mol C H O 1 mol C H O 180.15 g C H O3.077 g C H O

138.12 g C H O 1 mol C H O 1 mol C H O

= 4.01333 g C9H8O4 (unrounded)

Percent yield = ActualYield x 100%Theoretical Yield

= 3.281 g x 100%

4.01333 g

= 81.7526 = 81.75% yield

3-31

3.108 Plan: Determine the molecular formula and the molar mass of each of the compounds. From the amount of nitrogen present and the molar mass, the percent nitrogen may be determined. The moles need to be determined for part (b). Solution: a) To find mass percent of nitrogen, first determine molecular formula, then the molar mass of each compound. Mass percent is then calculated from the mass of nitrogen in the compound divided by the molar mass of the compound, and multiply by 100%. Urea: CH4N2O, M = 60.06 g/mol

% N = %100 ONCH g/mol 60.06

N) g/mol 01.14(2

24

= 46.6533 = 46.65% N in urea

Arginine: C6H15N4O2, M = 175.22 g/mol

% N = %100 ONHC g/mol 175.22

N) g/mol 01.14(4

24156

= 31.98265 = 31.98% N in arginine

Ornithine: C5H13N2O2, M = 133.17 g/mol

% N = %100 ONHC g/mol 133.17

N) g/mol 01.14(2

22135

= 21.04077 = 21.04% N in ornithine

b) Grams of N =

( ) 5 13 2 2 4 25 13 2 25 13 2 2 5 13 2 2 4 2

1 mol C H N O 1 mol CH N O 2 mol N 14.01 g N135.2 g C H N O133.17 g C H N O 1 mol C H N O 1 mol CH N O 1 mol N

= 28.447 = 28.45 g N 3.109 Plan: Write and balance the chemical reaction. Use the mole ratio to find the amount of product that should be

produced and take 66% of that amount to obtain the actual yield. Solution: 2 NO(g) + O2(g) 2 NO2(g)

With 6 molecules of NO and 3 molecules of O2 reacting, 6 molecules of NO2 can be produced. If the reaction only has a 66% yield, then (0.66)(6) = 4 molecules of NO2 will be produced. Circle A shows the formation of 4 molecules of NO2. Circle B also shows the formation of 4 molecules of NO2 but also has 2 unreacted molecules of NO and 1 unreacted molecule of O2. Since neither reactant is limiting, there will be no unreacted reactant left after reaction. 3.110 First, balance the given equation: 8 Zn(s) + S8(s) 8 ZnS(s) The reactions with oxygen are: 2 Zn(s) + O2(g) 2 ZnO(s) S8(s) + 8 O2(g) 8 SO2(g) a) Since the amounts of both starting materials are given, we must first find the limiting reagent:

Moles of ZnS from Zn = ( ) 1 mol Zn 8 mol ZnS83.2 g Zn 65.41 g Zn 8 mol Zn

= 1.27198 mol ZnS (unrounded)

Moles of ZnS from S8 = ( ) 888 8

1 mol S 8 mol ZnS52.4 g S256.56 g S 1 mol S

= 1.6339 mol ZnS (unrounded)

The zinc will produce less zinc sulfide, thus, zinc is the limiting reactant and will first be used to determine the theoretical yield and then the percent yield.

Theoretical yield = ( ) 1 mol Zn 8 mol ZnS 97.48 g ZnS83.2 g Zn 65.41 g Zn 8 mol Zn 1 mol ZnS

= 123.9923 g ZnS (unrounded)

Percent yield = ActualYield x 100%Theoretical Yield

= 104.4 g x 100%

123.9923 g

= 84.1988 = 84.2% yield

3-32

b) The theoretical yield indicates that 84.2% of the zinc produced zinc sulfide so (100 84.2)%= 15.8% of the zinc became zinc oxide. This allows the calcula